Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$

Let $\; I = \displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$

$\left[\text{Note: Integration by parts: } \displaystyle \int u \cdot v \; dx = u \displaystyle \int v \; dx - \displaystyle \int \left\{\dfrac{du}{dx} \times \displaystyle \int v \; dx \right\} dx \right]$

$\left[ \text{Here: } u = \log \left(\sin x\right), \; v = \cos 2x \right]$

$\begin{aligned} \therefore \; I & = \log \left(\sin x\right) \int \cos 2x \; dx - \int \left\{\dfrac{d}{dx} \left[\log \left(\sin x\right)\right] \times \int \cos 2x \; dx \right\} dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1}{\sin x} \times \cos x \times \dfrac{\sin 2x}{2} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{\cos x \times 2 \times \sin x \times \cos x}{2 \sin x} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \cos^2 x \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1 + \cos 2x}{2} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{1}{2} \int dx - \dfrac{1}{2} \int \cos 2x \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{x}{2} - \dfrac{\sin 2x}{4} + c \end{aligned}$

Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $I = \displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $x = \tan \theta$

Then, $\;$ $dx = \sec^2 \theta \; d\theta$

$\begin{aligned} \therefore \; I & = \int \sin^{-1} \left(\dfrac{2 \tan \theta}{1 + \tan^2 \theta}\right) \sec^2 \theta \; d\theta \\\\ & = \int \sin^{-1} \left[\sin \left(2 \theta\right)\right] \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin \left(2A\right) = \dfrac{2 \tan A}{1 + \tan^2 A}\right] \\\\ & = \int 2 \theta \; \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin^{-1} \left[\sin \left(A\right)\right] = A\right] \\\\ & = 2 \left\{\theta \int \sec^2 \theta \; d\theta - \int \left[\dfrac{d}{d \theta} \left(\theta\right) \times \int \sec^2 d \theta\right] d \theta \right\} \\\\ & \left[\text{Note: Integration by parts: } \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\ & = 2 \left\{\theta \; \tan \theta - \int \tan \theta \; d\theta \right\} + c \;\;\; \left[\text{Note: } \int \sec^2 x \; dx = \tan x + c\right] \\\\ & = 2 \; \theta \; \tan \theta - 2\log \left|\sec \theta\right| + c \;\;\; \left[\text{Note: } \int \tan x \; dx = \log \left|\sec x\right| + c\right] \\\\ & = 2 \; \theta \; \tan \theta - \log \left|\sec \theta\right|^2 + c \\\\ & = 2 \; \theta \; \tan \theta - \log \left|1 + \tan^2 \theta\right| + c \;\;\; \left[\text{Note: } 1 + \tan^2 x = \sec^2 x\right] \end{aligned}$

$\because \;$ $x = \tan \theta$ $\implies$ $\theta = \tan^{-1}x$

$\therefore \;$ $I = 2 \; x \; \tan^{-1}x - \log \left|1 + x^2\right| + c$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{1 + \sin x}{\sin x \left(1 + \cos x\right)} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{1 + \sin x}{\sin x \left(1 + \cos x\right)} \; dx \\\\ & = \int \dfrac{dx}{\sin x \left(1 + \cos x\right)} + \int \dfrac{dx}{1 + \cos x} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \text{Let } I_1 & = \int \dfrac{dx}{\sin x \left(1 + \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\sin^2 x \left(1 + \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos^2 x\right) \left(1 + \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos x\right) \left(1 + \cos x\right)^2} \;\;\; \cdots \; (2) \end{aligned}$

Let $\cos x = u$ $\;\;\; \cdots \; (3)$

Differentiating equation $(3)$ gives

$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (3a)$

$\therefore$ $\;$ In view of equations $(3)$ and $(3a)$, equation $(2)$ becomes

$I_1 = \displaystyle \int \dfrac{- du}{\left(1 - u\right) \left(1 + u\right)^2}$ $\;\;\; \cdots \; (4)$

Let $\dfrac{-1}{\left(1 - u\right) \left(1 + u\right)^2} = \dfrac{A}{1 - u} + \dfrac{B}{1 + u} + \dfrac{C}{\left(1 + u\right)^2}$ $\;\;\; \cdots \; (5)$

$\begin{aligned} \implies \; -1 & = A \left(1 + u\right)^2 + B \left(1 - u\right) \left(1 + u\right) + C \left(1 - u\right) \\\\ & = u^2 A + 2 u A + B - u^2 B + C - u C \\\\ & = u^2 \left(A - B\right) + u \left(2 A - C\right) + \left(A + B + C\right) \end{aligned}$

Comparing the coefficients of the $u^2$ term gives

$A - B = 0$ $\implies$ $A = B$ $\;\;\; \cdots \; (5a)$

Comparing the coefficients of the $u$ term gives

$2 A - C = 0$ $\implies$ $C = 2 A$ $\;\;\; \cdots \; (5b)$

Comparing the constant term gives

$-1 = A + B + C$ $\implies$ $-1 = A + A + 2 A$ $\;\;\;$ [by equations $(5a)$ and $(5b)$]

$\implies$ $4 A = -1$ $\implies$ $A = \dfrac{-1}{4}$

$\therefore$ $\;$ From equation $(5a)$, $B = \dfrac{-1}{4}$

From equation $(5b)$, $C = 2 \times \left(\dfrac{-1}{4}\right) = \dfrac{-1}{2}$

Substituting the values of A, B and C in equation $(5)$ gives

$\dfrac{-1}{\left(1 - u\right) \left(1 + u\right)^2} = \dfrac{-1}{4 \left(1 - u\right)} - \dfrac{1}{4 \left(1 + u\right)} - \dfrac{1}{2 \left(1 + u\right)^2}$ $\;\;\; \cdots \; (6)$

$\therefore$ $\;$ We have from equations $(4)$ and $(6)$

$\begin{aligned} I_1 & = \dfrac{-1}{4} \int \dfrac{du}{1 - u} - \dfrac{1}{4} \int \dfrac{du}{1 + u} - \dfrac{1}{2} \int \dfrac{du}{\left(1 + u\right)^2} \\\\ & = \dfrac{1}{4} \log \left|1 - u\right| - \dfrac{1}{4} \log \left|u + 1\right| + \dfrac{1}{2 \left(1 + u\right)} + c_1 \\\\ & = \dfrac{1}{4} \log \left|\dfrac{1 - u}{u + 1}\right| + \dfrac{1}{2 \left(1 + u\right)} + c_1 \\\\ & = \dfrac{1}{4} \log \left|\dfrac{1 - \cos x}{\cos x + 1}\right| + \dfrac{1}{2 \left(1 + \cos x\right)} + c_1 \;\;\; \left[\text{by equation } (3)\right] \\\\ & = \dfrac{1}{4} \log \left|\dfrac{2 \sin^2 \left(\dfrac{x}{2}\right)}{2 \cos^2 \left(\dfrac{x}{2}\right)}\right| + \dfrac{1}{2 \times 2 \cos^2 \left(\dfrac{x}{2}\right)} + c_1 \\\\ & = \dfrac{1}{4} \log \left|\tan^2 \left(\dfrac{x}{2}\right)\right| + \dfrac{1}{4} \sec^2 \left(\dfrac{x}{2}\right) + c_1 \;\;\; \cdots \; (7) \end{aligned}$

$\begin{aligned} \text{Let } I_2 & = \int \dfrac{dx}{1 + \cos x} \\\\ & = \int \dfrac{dx}{2 \cos^2 \left(\dfrac{x}{2}\right)} \\\\ & = \dfrac{1}{2} \int \sec^2 \left(\dfrac{x}{2}\right) \; dx \\\\ & = \dfrac{1}{2} \tan \left(\dfrac{x}{2}\right) \times 2 + c_2 \\\\ & = \tan \left(\dfrac{x}{2}\right) + c_2 \;\;\; \cdots \; (8) \end{aligned}$

$\therefore$ $\;$ In view of equations $(7)$ and $(8)$, equation $(1)$ becomes

$I = \dfrac{1}{4} \log \left|\tan^2 \left(\dfrac{x}{2}\right)\right| + \dfrac{1}{4} \sec^2 \left(\dfrac{x}{2}\right) + \tan \left(\dfrac{x}{2}\right) + c$

where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{\sin x \sqrt{\cos^3 x}}$

$\begin{aligned} \text{Let } I & = \int \dfrac{dx}{\sin x \sqrt{\cos^3 x}} \\\\ & = \int \dfrac{\sin x \; dx}{\sin^2 x \sqrt{\cos^3 x}} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos^2 x\right) \sqrt{\cos^3 x}} \;\;\; \cdots \; (1) \end{aligned}$

Let $\cos x = u$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (2a)$

$\therefore$ $\;$ In view of equations $(2)$ and $(2a)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \dfrac{- du}{\left(1 - u^2\right) \sqrt{u^3}} \\\\ & = \int \dfrac{- du}{u \sqrt{u} \left(1 + u\right) \left(1 - u\right)} \;\;\; \cdots \; (3) \end{aligned}$

Let $\sqrt{u} = v$ $\;\;\; \cdots \; (4)$

Differentiating equation $(4)$ gives

$\dfrac{1}{2 \sqrt{u}} \; du = dv$ $\implies$ $\dfrac{du}{\sqrt{u}} = 2 \; dv$ $\;\;\; \cdots \; (4a)$

Also from equation $(4)$, $u = v^2$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ In view of equations $(4)$, $(4a)$ and $(4b)$, equation $(3)$ becomes

$I = - 2 \displaystyle \int \dfrac{dv}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)}$ $\;\;\; \cdots \; (5)$

Let $v^2 = t$ (change of variable)

Then, $\dfrac{1}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} = \dfrac{1}{t \left(1 + t\right) \left(1 - t\right)}$ $\;\;\; \cdots \; (6)$

Let $\dfrac{1}{t \left(1 + t\right) \left(1 - t\right)} = \dfrac{A}{t} + \dfrac{B}{1 + t} + \dfrac{C}{1 - t}$ $\;\;\; \cdots \; (7)$

$\implies$ $1 = A \left(1 + t\right) \left(1 - t\right) + B \; t \left(1 - t\right) + C \; t \left(1 + t\right)$

$\implies$ $1 = A - At^2 + Bt - Bt^2 + Ct + Ct^2$

$\implies$ $1 = t^2 \left(- A - B + C\right) + t \left(B + C\right) + A$

Comparing the constant term gives

$A = 1$ $\;\;\; \cdots \; (7a)$

Comparing the coefficients of the $t$ term gives

$B + C = 0$ $\;\;\; \cdots \; (7b)$

Comparing the coefficients of the $t^2$ term gives

$- A - B + C = 0$ $\implies$ $- B + C = 1$ $\;\;\; \cdots \; (7c)$ $\;\;\;$ [by equation $(7a)$]

Adding equations $(7b)$ and $(7c)$ gives

$2 C = 1$ $\implies$ $C = \dfrac{1}{2}$

$\therefore$ $\;$ From equation $(7b)$, $B = - C = - \dfrac{1}{2}$

Substituting the values of A, B and C in equation $(7)$ gives

$\dfrac{1}{t \left(1 + t\right) \left(1 - t\right)} = \dfrac{1}{t} - \dfrac{1}{2 \left(1 + t\right)} - \dfrac{1}{2 \left(1 - t\right)}$ $\;\;\; \cdots \; (8)$

Since $v^2 = t$, equation $(8)$ becomes

$\dfrac{1}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} = \dfrac{1}{v^2} - \dfrac{1}{2 \left(1 + v^2\right)} + \dfrac{1}{2 \left(v^2 - 1\right)}$

$\begin{aligned} \therefore \; \int \dfrac{dv}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} & = \int \dfrac{dv}{v^2} - \dfrac{1}{2} \int \dfrac{dv}{v^2 + 1} + \dfrac{1}{2} \int \dfrac{dv}{v^2 -1} \\\\ & = \dfrac{-1}{v} - \dfrac{1}{2} \tan^{-1} \left(v\right) + \dfrac{1}{2} \int \dfrac{dv}{\left(v + 1\right) \left(v - 1\right)} \;\;\; \cdots \; (9) \end{aligned}$

Let $\dfrac{1}{\left(v + 1\right) \left(v - 1\right)} = \dfrac{P}{v + 1} + \dfrac{Q}{v -1}$ $\;\;\; \cdots \; (10)$

$\implies$ $1 = P \left(v - 1\right) + Q \left(v + 1\right)$

$\implies$ $1 = v \left(P + Q\right) + \left(Q - P\right)$

Comparing the coefficients of the $v$ term gives

$P + Q = 0$ $\implies$ $P = - Q$ $\;\;\; \cdots \; (10a)$

Comparing the constant term gives

$Q - P = 1$ $\implies$ $2 Q = 1$ $\implies$ $Q = \dfrac{1}{2}$ $\;\;\; \cdots \; (10b)$ [by equation $(10a)$]

$\therefore$ $\;$ From equation $(10a)$, $P = \dfrac{-1}{2}$

Substituting the values of P and Q in equation $(10)$ gives

$\dfrac{1}{\left(v + 1\right) \left(v - 1\right)} = \dfrac{-1}{2 \left(v + 1\right)} + \dfrac{1}{2 \left(v - 1\right)}$

$\begin{aligned} \therefore \; \dfrac{1}{2} \int \dfrac{dv}{\left(v + 1\right) \left(v - 1\right)} & = \dfrac{1}{2} \left[\dfrac{-1}{2} \int \dfrac{dv}{v + 1} + \dfrac{1}{2} \int \dfrac{dv}{v - 1}\right] \\\\ & = \dfrac{-1}{4} \log \left|v + 1\right| + \dfrac{1}{4} \log \left|v - 1\right| + c_1 \\\\ & = \dfrac{1}{4} \log \left|\dfrac{v - 1}{v + 1}\right| + c_1 \;\;\; \cdots \; (11) \end{aligned}$

$\therefore$ $\;$ In view of equation $(11)$, equation $(9)$ becomes

$\displaystyle \int \dfrac{dv}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} = \dfrac{-1}{v} - \dfrac{1}{2} \tan^{-1} \left(v\right) + \dfrac{1}{4} \log \left|\dfrac{v - 1}{v + 1}\right| + c_1$ $\;\;\; \cdots \; (12)$

$\therefore$ $\;$ In view of equation $(12)$, equation $(5)$ becomes

$\begin{aligned} I & = \dfrac{2}{v} + \tan^{-1} \left(v\right) - \dfrac{1}{2} \log \left|\dfrac{v - 1}{v + 1}\right| + c \;\;\; \left[\text{where } c =- 2 c_1\right] \\\\ & = \dfrac{2}{\sqrt{u}} + \tan^{-1} \left(\sqrt{u}\right) - \dfrac{1}{2} \log \left|\dfrac{\sqrt{u} - 1}{\sqrt{u} + 1}\right| + c \;\;\; \left[\text{by equation } (4)\right] \\\\ & = \dfrac{2}{\sqrt{\cos x}} + \tan^{-1} \left(\sqrt{\cos x}\right) - \dfrac{1}{2} \log \left|\dfrac{\sqrt{\cos x} -1 }{\sqrt{\cos x} + 1}\right| + c \;\;\; \left[\text{by equation } (2)\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \left(x - 5\right) \sqrt{x^2 + x} \; dx$

Let $I = \displaystyle \int \left(x - 5\right) \sqrt{x^2 + x} \; dx$ $\;\;\; \cdots \; (1)$

Let $x - 5 = M \; \dfrac{d}{dx} \left(x^2 + x\right) + N$ $\;\;\; \cdots \; (2)$

i.e. $x - 5 = M \left(2 x + 1\right) + N$

i.e. $x - 5 = 2 M x + \left(M + N\right)$

Comparing the coefficients of the $x$ term gives

$1 = 2 M$ $\implies$ $M = \dfrac{1}{2}$ $\;\;\; \cdots \; (3a)$

Comparing the constant term gives

$- 5 = M + N$

$\implies$ $N = - M - 5 = \dfrac{-1}{2} - 5 = \dfrac{-11}{2}$ $\;\;\; \cdots \; (3b)$ $\;\;\; $ [by equation $(3a)$]

In view of equations $(3a)$ and $(3b)$, equation $(2)$ can be written as

$x - 5 = \dfrac{1}{2} \left(2 x + 1\right) - \dfrac{11}{2}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \left[\dfrac{1}{2} \left(2x + 1\right) - \dfrac{11}{2}\right] \sqrt{x^2 + x} \; dx \\\\ & = \dfrac{1}{2} \int \left(2 x + 1\right) \sqrt{x^2 + x} \; dx - \dfrac{11}{2} \int \sqrt{x^2 + x} \; dx \;\;\; \cdots \; (5) \end{aligned}$

Let $I_1 = \dfrac{1}{2} \displaystyle \int \left(2 x + 1\right) \sqrt{x^2 + x} \; dx$ $\;\;\; \cdots \; (6)$

Let $x^2 + x = t$ $\;\;\; \cdots \; (6a)$

Differentiating equation $(6a)$ gives

$\left(2 x + 1\right) \; dx = dt$ $\;\;\; \cdots \; (6b)$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(6)$ becomes

$\begin{aligned} I_1 & = \dfrac{1}{2} \int \sqrt{t} \; dt \\\\ & = \dfrac{1}{2} \times t^{3/2} \times \dfrac{2}{3} + c_1 \\\\ & = \dfrac{1}{3} t^{3/2} + c_1 \\\\ & = \dfrac{1}{3} \left(x^2 + x\right)^{3/2} + c_1 \;\;\; \cdots \; (7) \;\;\; \left[\text{from equation }(6a)\right] \end{aligned}$

$\begin{aligned} \text{Let } I_2 & = - \dfrac{11}{2} \int \left(2x + 1\right) \sqrt{x^2 + x} \; dx \\\\ & = - \dfrac{11}{2} \int \sqrt{\left(x^2 + x + \dfrac{1}{4}\right) - \dfrac{1}{4}} \; dx \\\\ & = - \dfrac{11}{2} \int \sqrt{\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2} \; dx \\\\ & \left[\text{Note: } \int \sqrt{x^2 - a^2} \; dx = \dfrac{x}{2} \sqrt{x^2 - a^2} - \dfrac{a^2}{2} \log \left|x + \sqrt{x^2 - a^2}\right| + c\right] \\\\ \therefore \; I_2 & = \dfrac{-11}{2} \left(\dfrac{x + \dfrac{1}{2}}{2}\right) \sqrt{\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2} \\ & \hspace{5em} + \dfrac{11}{2} \times \dfrac{\left(1/2\right)^2}{2} \log \left|\left(x + \dfrac{1}{2}\right) + \sqrt{\left(x - \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2}\right| + c_2 \\\\ & = \dfrac{-11}{2} \left[\left(\dfrac{2x + 1}{4}\right) \sqrt{x^2 + x} - \dfrac{1}{8} \log \left|\dfrac{2x + 1}{2} + \sqrt{x^2 + x}\right|\right] + c_2 \\\\ & = \dfrac{-11}{8} \left(2 x + 1\right) \sqrt{x^2 + x} + \dfrac{11}{16} \log \left|\dfrac{2x + 1}{2} + \sqrt{x^2 + x}\right| + c_2 \;\;\; \cdots \; (8) \end{aligned}$

$\therefore$ $\;$ In view of equations $(7)$ and $(8)$, equation $(5)$ becomes

$I = \dfrac{1}{3} \left(x^2 + x\right)^{3/2} - \dfrac{11}{8} \left(2x + 1\right) \sqrt{x^2 + x} + \dfrac{11}{16} \log \left|\dfrac{2x + 1}{2} + \sqrt{x^2 + x}\right| + c$

where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{\log x - 1}{\left(\log x\right)^2} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{\log x - 1}{\left(\log x\right)^2} \; dx \\\\ & = \int \left[\dfrac{1}{\log x} - \dfrac{1}{\left(\log x\right)^2}\right] \; dx \;\;\; \cdots \; (1) \end{aligned}$

Let $\log x = t$ $\;\;\; \cdots (2a)$

$\implies$ $x = e^t$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2a)$ gives

$\dfrac{1}{x} \; dx = dt$ $\implies$ $dx = x \; dt$ $\implies$ $dx = e^t \; dt$ $\;\;\;$ [by equation $(2a)$] $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes

$I = \displaystyle \int \left[e^t \left(\dfrac{1}{t} - \dfrac{1}{t^2}\right)\right] \; dt$ $\;\;\; \cdots \; (3)$

If $f\left(t\right) = \dfrac{1}{t}$, then $f'\left(t\right) = \dfrac{-1}{t^2}$

$\therefore$ $\;$ Equation $(3)$ can be rewritten as

$I = \displaystyle \int e^t \left[\dfrac{1}{t} + \dfrac{d}{dt} \left(\dfrac{1}{t}\right)\right] \; dt$

i.e. $I = \dfrac{e^t}{t} + c = \dfrac{e^{\log x}}{\log x} + c$ $\;\;\;$ [by equation $(2a)$]

i.e. $I = \dfrac{x}{\log x} + c$

$\left[\text{Note: }\displaystyle \int e^x \left[f\left(x\right) + f'\left(x\right)\right] \; dx = e^x \; f\left(x\right) + c\right]$

Indefinite Integration

Evaluate $\displaystyle \int \log \left(x + \sqrt{x^2 + a^2}\right) \; dx$

Let $I = \displaystyle \int \log \left(x + \sqrt{x^2 + a^2}\right) \; dx$ $\;\;\; \cdots \; (1)$

Let $x = a \; \tan \theta$ $\;\;\; \cdots \; (2a)$

Differentiating equation $(2a)$ gives

$dx = a \; \sec^2 \theta \; d\theta$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \log \left(a \; \tan \theta + \sqrt{a^2 \; \tan^2 \theta + a^2}\right) a \; \sec^2 \theta \; d\theta \\\\ & = a \int \sec^2 \theta \; \log \left(a \; \tan \theta + a \; \sec \theta\right) \; d\theta \\\\ & \left[\begin{aligned} \text{Note: } & \int u \; v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \\\\ & \text{Here } u = \log \left(a \; \tan \theta + a \; \sec \theta\right), \;\; v = \sec^2 \theta \end{aligned}\right] \\\\ & = a \log \left|a \tan \theta + a \sec \theta\right| \int \sec^2 \theta \; d\theta \\ & \hspace{2em} - a \int \left[\int \sec^2 \theta \; d\theta \times \dfrac{d}{d\theta} \left[\log \left(a \tan \theta + a \sec \theta\right)\right]\right] \; d \theta \\\\ & = a \left\{\tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - \int \dfrac{\tan \theta \left(a \sec^2 \theta + a \sec \theta \tan \theta\right)}{a \left(\tan \theta + \sec \theta\right)} \; d \theta \right\} \\\\ & = a \left\{\tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - \int \dfrac{a \tan \theta \sec \theta \left(\sec \theta + \tan \theta\right)}{a \left(\tan \theta + \sec \theta\right)} \; d\theta \right\} \\\\ & = a \tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - a \int \tan \theta \; \sec \theta \; d\theta \;\;\; \cdots \; (3) \end{aligned}$

$\begin{aligned} \text{Consider } \int \tan \theta \; \sec \theta \; d\theta & = \int \dfrac{\sin \theta}{\cos \theta} \times \dfrac{1}{\cos \theta} \; d \theta \\\\ & = \int \dfrac{\sin \theta}{\cos^2 \theta} \; d \theta \;\;\; \cdots \; (4) \end{aligned}$

Let $\cos \theta = t$ $\;\;\; \cdots \; (5a)$

Differentiating equation $(5a)$ gives

$- \sin \theta \; d\theta = dt$ $\implies$ $\sin \theta \; d\theta = - dt$ $\;\;\; \cdots \; (5b)$

In view of equations $(5a)$ and $(5b)$, equation $(4)$ becomes

$\begin{aligned} \int \tan \theta \; \sec \theta \; d\theta & = \int \dfrac{- dt}{t^2} \\\\ & = \dfrac{1}{t} + c_1 \\\\ & = \dfrac{1}{\cos \theta} + c_1 \;\;\; \left[\text{from equation }(5a)\right] \;\;\; \cdots \; (6) \end{aligned}$

$\therefore$ $\;$ In view of equation $(6)$, equation $(3)$ becomes

$I = a \; \tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - \dfrac{a}{\cos \theta} + c$ $\;\;\; \cdots \; (7)$

where $c = -a \; c_1$

Now, from equation $(2a)$,

$\tan \theta = \dfrac{x}{a}$ $\;\;\; \cdots \; (8a)$

$\sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + \dfrac{x^2}{a^2}} = \dfrac{\sqrt{x^2 + a^2}}{a}$ $\;\;\; \cdots \; (8b)$

Substituting equations $(8a)$ and $(8b)$ in equation $(7)$ gives

$I = a \times \dfrac{x}{a} \times \log \left|a \times \dfrac{x}{a} + a \times \dfrac{\sqrt{x^2 + a^2}}{a}\right| - a \times \dfrac{\sqrt{x^2 + a^2}}{a} + c$

i.e. $I = x \; \log \left|x + \sqrt{x^2 + a^2}\right| - \sqrt{x^2 + a^2} + c$

Indefinite Integration

Evaluate $\displaystyle \int \tan^{-1} \left(\sqrt{\dfrac{1 - x}{1 + x}}\right) \; dx$

$\begin{aligned} \text{Let } I & = \int \tan^{-1} \left(\sqrt{\dfrac{1 - x}{1 + x}}\right) \; dx \\\\ & = \int \tan^{-1} \left(\dfrac{1 - x}{\sqrt{1 - x^2}}\right) \; dx \;\;\; \cdots \; (1) \end{aligned}$

Let $x = \cos \theta$ $\;\;\; \cdots \; (2a)$

Differentiating equation $(2a)$ gives

$dx = - \sin \theta \; d\theta$ $\;\;\; \cdots \; (2b)$

$\begin{aligned} \text{Now, } \dfrac{1 - x}{\sqrt{1 - x^2}} & = \dfrac{1 - \cos \theta}{\sqrt{1 - \cos^2 \theta}} \;\;\; \left[\text{by equation }(2a)\right] \\\\ & = \dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{\sin \theta} \;\;\; \left[\text{Note: }1 - \cos 2 \theta = 2 \sin^2 \theta\right] \\\\ & = \dfrac{2 \sin^2 \left(\dfrac{\theta}{2}\right)}{2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)} \;\;\; \left[\text{Note: } \sin 2 \theta = 2 \sin \theta \cos \theta\right] \\\\ & = \dfrac{\sin \left(\dfrac{\theta}{2}\right)}{\cos \left(\dfrac{\theta}{2}\right)} = \tan \left(\dfrac{\theta}{2}\right) \;\;\; \cdots \; (2c) \end{aligned}$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \tan^{-1} \left[\tan \left(\dfrac{\theta}{2}\right)\right] \left(- \sin \theta\right) \; d\theta \\\\ & = - \int \dfrac{\theta}{2} \; \sin \theta \; d\theta \\\\ & \left[\begin{aligned} \text{Note: } & \int u \; v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \\\\ & \text{Here } u = \theta, \;\; v = \sin \theta \end{aligned}\right] \\\\ & = \dfrac{-1}{2} \left\{\theta \int \sin \theta \; d \theta - \int \left[\int \sin \theta \; d \theta \times \dfrac{d}{d\theta} \left(\theta\right)\right] \; d\theta \right\} \\\\ & = \dfrac{-1}{2} \left\{- \theta \cos \theta + \int \cos \theta \; d \theta \right\} \\\\ & = \dfrac{\theta \; \cos \theta}{2} - \dfrac{\sin \theta}{2} + c \;\;\; \cdots \; (3) \end{aligned}$

From equation $(2a)$,

$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$ $\;\;\; \cdots \; (4a)$

and $\theta = \cos^{-1}\left(x\right)$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ In view of equations $(4a)$ and $(4b)$, equation $(3)$ becomes

$I = \dfrac{x \; \cos^{-1}x}{2} - \dfrac{\sqrt{1 - x^2}}{2} + c$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{\sin x - \sin 2x}$

$\begin{aligned} \text{Let } I & = \int \dfrac{dx}{\sin x - \sin 2x} \\\\ & = \int \dfrac{dx}{\sin x - 2 \sin x \cos x} \\\\ & = \int \dfrac{dx}{\sin x \left(1 - 2 \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\sin^2 x \left(1 - 2 \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos^2 x\right) \left(1 - 2 \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 + \cos x\right) \left(1 - \cos x\right) \left(1 - 2 \cos x\right)} \;\;\; \cdots \; (1) \end{aligned}$

Let $\cos x = u$ $\;\;\; \cdots (2a)$

Differentiating equation $(2a)$ gives

$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$I = \displaystyle \int \dfrac{- du}{\left(1 + u\right) \left(1 - u\right) \left(1 - 2u\right)}$ $\;\;\; \cdots \; (3)$

Let $\dfrac{-1}{\left(1 + u\right) \left(1 - u\right) \left(1 - 2 u\right)} = \dfrac{A}{1 + u} + \dfrac{B}{1 - u} + \dfrac{C}{1 - 2 u}$ $\;\;\; \cdots \; (4)$

$\implies$ $-1 = A \left(1 - u\right) \left(1 - 2u\right) + B \left(1 + u\right)\left(1 - 2u\right) + C \left(1 + u\right) \left(1 - u\right)$

$\begin{aligned} \text{When } u = -1, \hspace{2em} & -1 = A \times 2 \times 3 & \text{i.e. } -1 = 6 A & \implies A = \dfrac{-1}{6} \\\\ \text{When } u = 1, \hspace{2em} & -1 = B \times 2 \times \left(-1\right) & \text{i.e. } 1 = 2 B & \implies B = \dfrac{1}{2} \\\\ \text{When } u = \dfrac{1}{2}, \hspace{2em} & -1 = C \times \dfrac{3}{2} \times \dfrac{1}{2} & \text{i.e. } -1 = \dfrac{3 \; C}{4} & \implies C = \dfrac{-4}{3} \end{aligned}$

Substituting the values of A, B and C in equation $(4)$ gives

$\dfrac{-1}{\left(1 + u\right) \left(1 - u\right) \left(1 - 2 u\right)} = \dfrac{-1}{6 \left(1 + u\right)} + \dfrac{1}{2 \left(1 - u\right)} + \dfrac{-4}{3 \left(1 - 2 u\right)}$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equation $(5)$, equation $(3)$ becomes

$\begin{aligned} I & = \dfrac{-1}{6} \int \dfrac{du}{1 + u} + \dfrac{1}{2} \int \dfrac{du}{1 - u} - \dfrac{4}{3} \int \dfrac{du}{1 - 2 u} \\\\ & = \dfrac{-1}{6} \log \left|1 + u\right| - \dfrac{1}{2} \log \left|1 - u\right| + \dfrac{2}{3} \log \left|1 - 2 u\right| + c \\\\ & = \dfrac{-1}{6} \log \left|1 + \cos x\right| - \dfrac{1}{2} \log \left|1 - \cos x\right| + \dfrac{2}{3} \log \left|1 - 2 \cos x\right| + c \\\\ & \;\;\; \left[\text{from equation } (2a)\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{\left(x + 1\right)^2 \left(x^2 + 1\right)}$

Let $I = \displaystyle \int \dfrac{dx}{\left(x + 1\right)^2 \left(x^2 + 1\right)}$ $\;\;\;\cdots \; (1)$

Let $\dfrac{1}{\left(x + 1\right)^2 \left(x^2 + 1\right)} = \dfrac{A}{x + 1} + \dfrac{B}{\left(x + 1\right)^2} + \dfrac{C x + D}{x^2 + 1}$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \text{i.e. } 1 & = A \left(x + 1\right) \left(x^2 + 1\right) + B \left(x^2 + 1\right) + \left(C x + D\right) \left(x + 1\right)^2 \\\\ & = A x^3 + A x^2 + A x + A + Bx^2 + B + C x^3 + 2 C x^2 + C x + D x^2 + 2 D x + D \\\\ & = x^3 \left(A + C\right) + x^2 \left(A + B + 2 C + D\right) + x \left(A + C + 2 D\right) + \left(A + B + D\right) \end{aligned}$

Comparing the constant term gives

$A + B + D = 1$ $\;\;\; \cdots \; (3a)$

Comparing the coefficients of the $x$ term gives

$A + C + 2 D = 0$ $\;\;\; \cdots \; (3b)$

Comparing the coefficients of the $x^2$ term gives

$A + B + 2 C + D = 0$ $\;\;\; \cdots \; (3c)$

Comparing the coefficients of the $x^3$ term gives

$A + C = 0$ $\implies$ $C = - A$ $\;\;\; \cdots \; (3d)$

Substituting equation $(3d)$ in equation $(3b)$ gives

$2 D = 0$ $\implies$ $D = 0$

Substituting the value of D in equation $(3a)$ gives

$A + B = 1$ $\;\;\; \cdots \; (4a)$

In view of equation $(3d)$ and $D = 0$, equation $(3c)$ becomes

$A + B -2 A = 0$ $\implies$ $B - A = 0$ $\;\;\; \cdots \; (4b)$

Adding equations $(4a)$ and $(4b)$ gives

$2 B = 1$ $\implies$ $B = \dfrac{1}{2}$

$\therefore$ From equation $(4b)$, $A = \dfrac{1}{2}$

$\therefore$ From equation $(3d)$, $C = \dfrac{-1}{2}$

Substituting the values of A, B, C and D in equation $(2)$ gives

$\dfrac{1}{\left(x + 1\right)^2 \left(x^2 + 1\right)} = \dfrac{1}{2 \left(x + 1\right)} + \dfrac{1}{2 \left(x + 1\right)^2} - \dfrac{x}{2 \left(x^2 + 1\right)}$ $\;\;\; \cdots \; (5)$

$\therefore$ In view of equation $(5)$, equation $(1)$ becomes

$I = \dfrac{1}{2} \displaystyle \int \dfrac{dx}{x + 1} + \dfrac{1}{2} \displaystyle \int \dfrac{dx}{\left(x + 1\right)^2} - \dfrac{1}{2} \displaystyle \int \dfrac{x \; dx}{x^2 + 1}$

i.e. $I = \dfrac{1}{2} \log \left|x + 1\right| - \dfrac{1}{2 \left(x + 1\right)} - \dfrac{1}{2} \displaystyle \int \dfrac{x \; dx}{x^2 + 1}$ $\;\;\; \cdots \; (6)$

Consider $\displaystyle \int \dfrac{x \; dx}{x^2 + 1}$ $\;\;\; \cdots \; (7)$

Let $x^2 + 1 = t$ $\;\;\; \cdots \; (7a)$

Differentiating equation $(7a)$ gives

$2 x \; dx = dt$ $\implies$ $x \; dx = \dfrac{dt}{2}$ $\;\;\; \cdots \; (7b)$

$\therefore$ $\;$ In view of equations $(7a)$ and $(7b)$, equation $(7)$ becomes

$\begin{aligned} \displaystyle \int \dfrac{x \; dx}{x^2 + 1} & = \dfrac{1}{2} \displaystyle \int \dfrac{dt}{t} \\\\ & = \dfrac{1}{2} \log \left|t\right| + c \\\\ & = \dfrac{1}{2} \log \left|x^2 + 1\right| + c \;\;\; \cdots \; (8) \;\;\; \left[\text{from equation }(7a)\right] \end{aligned}$

$\therefore$ $\;$ We have from equations $(6)$ and $(8)$

$I = \dfrac{1}{2} \log \left|x + 1\right| - \dfrac{1}{2 \left(x + 1\right)} - \dfrac{1}{4} \log \left|x^2 + 1\right| + c$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{5 x}{\left(x + 1\right) \left(x^2 + 9\right)} \; dx$

Let $I = \displaystyle \int \dfrac{5 x}{\left(x + 1\right) \left(x^2 + 9\right)} \; dx$ $\;\;\; \cdots \; (1)$

Let $\dfrac{5x}{\left(x + 1\right) \left(x^2 + 9\right)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 9}$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \text{i.e. } \; 5x & = A \left(x^2 + 9\right) + \left(Bx + C\right) \left(x + 1\right) \\\\ & = A x^2 + 9 A + B x^2 + B x + C x + C \\\\ & = x^2 \left(A + B\right) + x \left(B + C\right) + \left(9 A + C\right) \end{aligned}$

Comparing the coefficient of $x^2$ term gives

$A + B = 0$ $\implies$ $B = - A$ $\;\;\; \cdots \; (3a)$

Comparing the coefficient of $x$ term gives

$B + C = 5$ $\implies$ $- A + C = 5$ $\;\;\; \cdots \; (3b)$ [from equation $(3a)$]

Comparing the constant term gives

$9 A + C = 0$ $\;\;\; \cdots \; (3c)$

Subtracting equations $(3b)$ and $(3c)$ gives

$10 A = - 5$ $\implies$ $A = \dfrac{-1}{2}$

$\therefore$ $\;$ We have from equation $(3a)$, $B = -A = \dfrac{1}{2}$

From equation $(3b)$, $C = 5 + A = 5 - \dfrac{1}{2} = \dfrac{9}{2}$

Substituting the values of A, B and C in equation $(2)$ gives

$\dfrac{5 x}{\left(x + 1\right) \left(x^2 + 9\right)} = \dfrac{-1}{2 \left(x + 1\right)} + \dfrac{\dfrac{1}{2}x + \dfrac{9}{2}}{x^2 + 9}$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes

$I = \dfrac{-1}{2} \displaystyle \int \dfrac{dx}{x + 1} + \dfrac{1}{2} \displaystyle \int \dfrac{x}{x^2 + 9} \; dx + \dfrac{9}{2} \displaystyle \int \dfrac{dx}{x^2 + 9}$ $\;\;\; \cdots \; (5)$

Now, $\displaystyle \int \dfrac{dx}{x + 1} = \log \left|x + 1\right| + c_1$ $\;\;\; \cdots \; (6a)$

Consider $\displaystyle \int \dfrac{x}{x^2 + 9} \; dx$

Let $x^2 + 9 = t$ $\;\;\; \cdots \; (7a)$

Differentiating equation $(7a)$ w.r.t x gives

$2 x \; dx = dt$ $\implies$ $x \; dx = \dfrac{dt}{2}$ $\;\;\; \cdots \; (7b)$

$\therefore$ $\;$ In view of equations $(7a)$ and $(7b)$ we have

$\begin{aligned} \int \dfrac{x \; dx}{x^2 + 9} & = \dfrac{1}{2} \int \dfrac{dt}{t} \\\\ & = \dfrac{1}{2} \log \left|t\right| + c_2 \\\\ & = \dfrac{1}{2} \log \left|x^2 + 9\right| + c_2 \;\;\; \left[\text{from equation (7a)}\right] \;\;\; \cdots (6b) \end{aligned}$

$\begin{aligned} \text{Consider } \int \dfrac{dx}{x^2 + 9} & = \int \dfrac{dx}{\left(x\right)^2 + \left(3\right)^2} \\\\ & = \dfrac{1}{3} \tan^{-1} \left(\dfrac{x}{3}\right) + c_3 \;\;\; \cdots \; (6c) \\\\ & \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c\right] \end{aligned}$

$\therefore$ $\;$ In view of equations $(6a)$, $(6b)$ and $(6c)$, equation $(5)$ becomes

$I = \dfrac{-1}{2} \log \left|x + 1\right| + \dfrac{1}{4} \log \left|x^2 + 9\right| + \dfrac{3}{2} \tan^{-1} \left(\dfrac{x}{3}\right) + c$

where $c = \dfrac{- c_1}{2} + \dfrac{c_2}{2} + \dfrac{9 c_3}{2}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^2 + x + 1}{\left(x + 1\right)^2 \left(x + 2\right)} \; dx$

Let $I = \displaystyle \int \dfrac{x^2 + x + 1}{\left(x + 1\right)^2 \left(x + 2\right)} \; dx$ $\;\;\; \cdots \; (1)$

Let $\dfrac{x^2 + x + 1}{\left(x + 1\right)^2 \left(x + 2\right)} = \dfrac{A}{\left(x + 1\right)^2} + \dfrac{B}{x + 1} + \dfrac{C}{x + 2}$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \implies x^2 + x + 1 & = A \left(x + 2\right) + B \left(x + 1\right) \left(x + 2\right) + C \left(x + 1\right)^2 \\\\ & = A x + 2 A + B x^2 + 3 B x + 2 B + C x^2 + 2 C x + C \\\\ & = x^2 \left(B + C\right) + x \left(A + 3 B + 2 C\right) + \left(2 A + 2 B + C\right) \end{aligned}$

Comparing the coefficient of $x^2$ term gives

$B + C = 1$ $\implies$ $B = 1 - C$ $\;\;\; \cdots \; (3a)$

Comparing the coefficient of $x$ term gives

$A + 3 B + 2 C = 1$ $\implies$ $A + 3 \left(1 - C\right) + 2 C = 1$ $\;\;\;$ [By equation $(3a)$]

$\implies$ $A - C = - 2$ $\;\;\; \cdots \; (3b)$

Comparing the constant term gives

$2 A + 2 B + C = 1$ $\implies$ $2 A + 2 \left(1 - C\right) + C = 1$ $\;\;\;$ [By equation $(3a)$]

$\implies$ $2 A - C = -1$ $\;\;\; \cdots \; (3c)$

Subtracting equations $(3a)$ and $(3b)$ gives $A = 1$

Substituting the value of A in equation $(3b)$ gives $C = A + 2 = 3$

Substituting the value of B in equation $(3a)$ gives $B = 1 - C = -2$

$\therefore$ $\;$ Substituting the values of A, B and C in equation $(2)$ gives

$\dfrac{x^2 + x + 1}{\left(x + 1\right)^2 \left(x + 2\right)} = \dfrac{1}{\left(x + 1\right)^2} - \dfrac{2}{x + 1} + \dfrac{3}{x + 2}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \dfrac{dx}{\left(x + 1\right)^2} - 2 \int \dfrac{dx}{x + 1} + 3 \int \dfrac{dx}{x + 2} \\\\ & = \dfrac{- 1}{x + 1} - 2 \log \left|x + 1\right| + 3 \log \left|x + 2\right| + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} \; dx$

Let $I = \displaystyle \int \dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} \; dx$ $\;\;\; \cdots \; (1)$

Let $x^2 = p$ (change of variable; not substitution)

Then, $\dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} = \dfrac{p}{\left(2 p + 1\right) \left(p - 1\right)}$ $\;\;\; \cdots \; (2)$

Let $\dfrac{p}{\left(2 p + 1\right) \left(p - 1\right)} = \dfrac{A}{2 p + 1} + \dfrac{B}{p - 1}$ $\;\;\; \cdots \; (3)$

$\implies$ $p = A \left(p - 1\right) + B \left(2 p + 1\right)$

When $p = \dfrac{-1}{2}$,

$\dfrac{-1}{2} = \dfrac{-3}{2}A$ $\implies$ $A = \dfrac{1}{3}$

When $p = 1$,

$1 = 3 B$ $\implies$ $B = \dfrac{1}{3}$

Substituting the values of A and B in equation $(3)$ gives

$\dfrac{p}{\left(2 p + 1\right) \left(p - 1\right)} = \dfrac{1/3}{2 p + 1} + \dfrac{1/3}{p - 1}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(2)$ becomes

$\dfrac{x^2}{\left(2 x^2 + 1\right) \left(x^2 - 1\right)} = \dfrac{1/3}{2 x^2 + 1} + \dfrac{1/3}{x^2 - 1}$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ We have from equations $(5)$ and $(1)$

$\begin{aligned} I & = \dfrac{1}{3} \int \dfrac{dx}{2 x^2 + 1} + \dfrac{1}{3} \int \dfrac{dx}{x^2 - 1} \\\\ & = \dfrac{1}{6} \int \dfrac{dx}{x^2 + \left(1/ \sqrt{2}\right)^2} + \dfrac{1}{3} \int \dfrac{dx}{x^2 - \left(1\right)^2} \\\\ & = \dfrac{1}{6} \times \sqrt{2} \tan^{-1} \left(x \sqrt{2}\right) + \dfrac{1}{3} \times \dfrac{1}{2} \log \left|\dfrac{x - 1}{x + 1}\right| + c \\\\ & = \dfrac{1}{3 \sqrt{2}} \tan^{-1} \left(x \sqrt{2}\right) + \dfrac{1}{6} \log \left|\dfrac{x - 1}{x + 1}\right| + c \end{aligned}$

$\left[\begin{aligned} \text{Note: } & \\\\ & \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c \\\\ & \int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a} \log \left|\dfrac{x - a}{x + a}\right| + c \end{aligned}\right]$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^3 - 6 x^2 + 10 x - 2}{x^2 - 5 x + 6} \; dx$

Let $I = \displaystyle \int \dfrac{x^3 - 6 x^2 + 10 x - 2}{x^2 - 5 x + 6} \; dx$ $\;\;\; \cdots \; (1)$

$\begin{array}{rll} x^2 - 5 x + 6 ) & x^3 - 6 x^2 + 10 x - 2 & (x - 1 \\ & \underline{x^3 - 5 x^2 + 6 x} & \\ & \hspace{6mm} - x^2 \; + 4 x \; - 2 & \\ & \hspace{6mm} \underline{- x^2 \; + 5 x \; - 6} & \\ & \hspace{15mm} - x \; + \; 4 & \end{array}$

$\therefore$ $\;$ $\dfrac{x^3 - 6 x^2 + 10 x - 2}{x^2 - 5 x + 6} = x - 1 + \dfrac{4 - x}{x^2 - 5 x + 6}$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$I = \displaystyle \int x \; dx - \displaystyle \int dx + \displaystyle \int \dfrac{4 - x}{x^2 - 5 x + 6} \; dx$ $\;\;\; \cdots \; (3)$

Now, $\displaystyle \int x \; dx = \dfrac{x^2}{2} + c_1$ $\;\;\; \cdots \; (4)$

$\displaystyle \int dx = x + c_2$ $\;\;\; \cdots (5)$

$\displaystyle \int \dfrac{4 - x}{x^2 - 5 x + 6} \; dx = \displaystyle \int \dfrac{4 - x}{\left(x - 2\right) \left(x - 3\right)} \; dx$ $\;\;\; \cdots \; (6)$

Let $\dfrac{4 - x}{\left(x - 2\right) \left(x - 3\right)} = \dfrac{A}{x - 2} + \dfrac{B}{x - 3}$ $\;\;\; \cdots \; (6a)$

i.e. $4 - x = A \left(x - 3\right) + B \left(x - 2\right)$

When $x = 2$, $2 = - A$ $\implies$ $A = -2$

When $x = 3$, $B = 1$

Substituting the values of A and B in equation $(6a)$ gives

$\dfrac{4 - x}{\left(x - 2\right) \left(x - 3\right)} = \dfrac{-2}{x - 2} + \dfrac{1}{x - 3}$ $\;\;\; \cdots \; (6b)$

$\therefore$ $\;$ In view of equation $(6b)$ equation $(6)$ becomes

$\begin{aligned} \int \dfrac{4 - x}{x^2 - 5 x + 6} \; dx & = - 2 \int \dfrac{dx}{x - 2} + \int \dfrac{dx}{x - 3} \\\\ & = -2 \log \left|x - 2\right| + \log \left|x - 3\right| + c_3 \;\;\; \cdots \; (6c) \end{aligned}$

$\therefore$ $\;$ In view of equations $(4)$, $(5)$ and $(6c)$, equation $(3)$ becomes

$I = \dfrac{x^2}{2} - x - 2 \log \left|x - 2\right| + \log \left|x - 3\right| + c$

where $c = c_1 - c_2 + c_3$

Indefinite Integration

Evaluate $\displaystyle \int 3^x \; \sin^2 x \; dx$

$\begin{aligned} \text{Let } I & = \int 3^x \; \sin^2 x \; dx \\\\ & = \int 3^x \left[\dfrac{1 - \cos \left(2x\right)}{2}\right] \; dx \;\;\; \left[\text{Note: }\sin^2 \theta = \dfrac{1 - \cos \left(2 \theta\right)}{2} \right] \\\\ & = \dfrac{1}{2} \int 3^x \; dx - \dfrac{1}{2} \int 3^x \; \cos \left(2 x\right) \; dx \\\\ & = \dfrac{1}{2} I_1 - \dfrac{1}{2} I_2 \;\;\; \cdots (1) \end{aligned}$

Now, $I_1 = \displaystyle \int 3^x \; dx = \dfrac{3^x}{\log 3} + c_1$ $\;\;\; \cdots (2)$

$\begin{aligned} \text{and } I_2 & = \int 3^x \; \cos \left(2 x\right) \; dx \\\\ & \left[\begin{aligned} \text{Note: } & \int u \; v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \\ & \text{Here } u = \cos \left(2x\right) \text{ and } v = 3^x \end{aligned} \right] \\\\ \therefore \; I_2 & = \cos \left(2x\right) \int 3^x \; dx - \int \left\{\int 3^x \; dx \times \dfrac{d}{dx} \left[\cos \left(2x\right)\right] \right\} \; dx \\\\ & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\log 3} \int 3^x \; \sin \left(2x\right) \; dx \\\\ & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\log 3} \left[\sin \left(2x\right) \int 3^x \; dx - \int \left\{\int 3^x \; dx \times \dfrac{d}{dx} \left[\sin \left(2x\right)\right] \right\} \; dx\right] \\\\ & \left[\text{Note: Here } u = \sin \left(2 x\right) \text{ and } v = 3^x\right] \\\\ i.e. \; I_2 & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\log 3} \left[\dfrac{3^x \; \sin \left(2 x\right)}{\log 3} - \dfrac{2}{\log 3} \int 3^x \; \cos \left(2 x\right) \; dx\right] + c'_2 \\\\ & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\left(\log 3\right)^2} \times 3^x \; \sin \left(2 x\right) - \dfrac{4}{\left(\log 3\right)^2} \int 3^x \; \cos \left(2 x\right) \; dx + c'_2 \\\\ & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\left(\log 3\right)^2} \times 3^x \; \sin \left(2 x\right) - \dfrac{4}{\left(\log 3\right)^2} I_2 + c'_2 \end{aligned}$

$\begin{aligned} i.e. \; \left[1 + \dfrac{4}{\left(\log 3\right)^2}\right] \; I_2 & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\left(\log 3\right)^2} \times 3^x \; \sin \left(2 x\right) + c'_2 \\\\ i.e. \; \left[\dfrac{\left(\log 3\right)^2 + 4}{\left(\log 3\right)^2}\right] \; I_2 & = \dfrac{3^x \; \cos \left(2x\right)}{\log 3} + \dfrac{2}{\left(\log 3\right)^2} \times 3^x \; \sin \left(2 x\right) + c'_2 \end{aligned}$

$\begin{aligned} i.e. \; I_2 & = \dfrac{\left(\log 3\right) \times 3^x \times \cos \left(2 x\right)}{\left(\log 3\right)^2 + 4} + \dfrac{2 \times 3^x \times \sin \left(2 x\right)}{\left(\log 3\right)^2 + 4} + c_2 \;\;\;\; \cdots (3) \end{aligned}$

where $c_2 = \left[\dfrac{\left(\log 3\right)^2}{\left(\log 3\right)^2 + 4}\right] \times c'_2$

Substituting equations $(2)$ and $(3)$ in equation $(1)$ gives

$I = \dfrac{3^x}{2 \; \log 3} - \dfrac{3^x}{2 \left[\left(\log 3\right)^2 + 4\right]} \left[\left(\log 3\right) \cos \left(2 x\right) + 2 \sin \left(2 x\right)\right] + c$

where $c = \dfrac{c_1}{2} - \dfrac{c_2}{2}$

Indefinite Integration

Evaluate $\displaystyle \int e^{-x/2} \; \cos^2 x \; dx$

Let $I = \displaystyle \int e^{-x/2} \; \cos^2 x \; dx$ $\;\;\; \cdots (1)$

Let $\dfrac{- x}{2} = p$ $\;\;\; \cdots (2a)$

Differentiating equation $(2a)$ $\;$ w.r.t x gives

$\dfrac{- 1}{2} \; dx = dp$ $\implies$ $dx = - 2 \; dp$ $\;\;\; \cdots (2b)$

From equation $(2a)$, we have $x = - 2 p$

$\therefore$ $\;$ $\cos x = \cos \left(-2 p\right) = \cos \left(2 p\right)$ $\;\;\; \cdots (2c)$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can becomes

$\begin{aligned} I & = - 2 \int e^p \; \cos^2 \left(2 p\right) \; dp \\\\ & = - 2 \int e^p \left(\dfrac{1 + \cos \left(4 p\right)}{2}\right) \; dp \;\;\; \left[\text{Note: } \cos^2 \theta = \dfrac{1 + \cos \left(2 \theta\right)}{2}\right] \\\\ & = - \int e^p \; dp - \int e^p \; \cos \left(4 p\right) \; dp \;\;\; \cdots (3) \end{aligned}$

$\left[\text{Note: } \displaystyle \int u \; v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(v\right) \right\} \; dx \right]$

Consider $I_1 = \displaystyle \int e^p \; \cos \left(4 p\right) \; dp$

Here $u = \cos \left(4 p\right)$ and $v = e^p$

$\begin{aligned} \therefore \; I_1 & = \cos \left(4 p\right) \int e^p \; dp - \int \left\{\int e^p \; dp \times \dfrac{d}{dp} \left[\cos \left(4 p\right)\right] \right\} \; dp \\\\ & = e^p \; \cos \left(4 p\right) + 4 \int e^p \; \sin \left(4 p\right) \; dp \\\\ & = e^p \; \cos \left(4 p\right) + 4 \left[\sin \left(4 p\right) \int e^p \; dp - \int \left\{\int e^p \; dp \times \dfrac{d}{dp} \left[\sin \left(4 p\right)\right] \right\} \; dp\right] \\\\ & \left[\text{Note: Here } u = \sin \left(4 p\right) \text{ and } v = e^p \right] \\\\ & = e^p \; \cos \left(4 p\right) + 4 \left[e^p \sin \left(4 p\right) - 4 \int e^p \; \cos \left(4 p\right) \; dp \right] \\\\ & = e^p \; \cos \left(4 p\right) + 4 \; e^p \; \sin \left(4 p\right) - 16 \int e^p \; \cos \left(4 p\right) \; dp \\\\ \text{i.e.} \; 17 \; I_1 & = e^p \; \cos \left(4 p\right) + 4 \; e^p \; \sin \left(4 p\right) + c_1 \\\\ \therefore \; I_1 & = \dfrac{1}{17} \left[e^p \; \cos \left(4 p\right) + 4 \; e^p \; \sin \left(4 p\right)\right] + \dfrac{c_1}{17} \;\;\; \cdots (4) \end{aligned}$

$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes

$\begin{aligned} I & = - e^p - \dfrac{1}{17} \left[e^p \; \cos \left(4 p\right) + 4 \; e^p \; \sin \left(4 p\right)\right] + c \;\;\; \left[\text{where } c = \dfrac{-c_1}{17}\right] \\\\ & = -e^{-x/2} - \dfrac{e^{-x/2}}{17} \left[\cos \left(\dfrac{-4 x}{2}\right) + 4 \sin \left(\dfrac{-4 x}{2}\right)\right] + c \;\;\; \left[\text{From equation (2a)}\right] \\\\ & = -e^{-x/2} - \dfrac{e^{-x/2}}{17} \left[\cos \left(2x\right) - 4 \sin \left(2x\right)\right] + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \left(3 x - 2\right) \sqrt{x^2 + x + 1} \;\; dx$

Let $I = \displaystyle \int \left(3 x - 2\right) \sqrt{x^2 + x + 1} \;\; dx$ $\;\;\; \cdots (1)$

$\begin{aligned} \text{Let } 3 x - 2 & = M \dfrac{d}{dx} \left(x^2 + x + 1\right) + N \\\\ & = M \left(2 x + 1\right) + N \;\;\; \cdots (2) \\\\ & = 2 \; M \; x + M + N \end{aligned}$

Comparing the coefficient of the x terms and the constant terms gives

$2 M = 3$ $\implies$ $M = \dfrac{3}{2}$

and $- 2 = M + N$ $\implies$ $N = - M - 2 = - \dfrac{3}{2} - 2 = \dfrac{- 7}{2}$

Substituting the values of M and N in equation (2) gives

$3 x - 2 = \dfrac{3}{2} \left(2 x + 1\right) - \dfrac{7}{2}$ $\;\;\; \cdots (2a)$

In view of equation (2a), equation (1) can be written as

$\begin{aligned} I & = \int \left[\dfrac{3}{2} \left(2 x + 1\right) - \dfrac{7}{2}\right] \sqrt{x^2 + x + 1} \; dx \\\\ & = \dfrac{3}{2} \int \left(2 x + 1\right) \sqrt{x^2 + x + 1} \; dx - \dfrac{7}{2} \int \sqrt{x^2 + x + 1} \; dx \\\\ & = \dfrac{3}{2} I_1 - \dfrac{7}{2} I_2 \;\;\; \cdots (3) \end{aligned}$

Consider $I_1 = \displaystyle \int \left(2 x + 1\right) \sqrt{x^2 + x + 1} \; dx$ $\;\;\; \cdots (4)$

Let $x^2 + x + 1 = u$ $\;\;\; \cdots (5a)$

Differentiating equation $(5a)$ gives

$\left(2 x + 1\right) \; dx = du$ $\;\;\; \cdots (5b)$

Substituting equations $(5a)$ and $(5b)$ in equation $(4)$ gives

$\begin{aligned} I_1 & = \int \sqrt{u} \; du \\\\ & = \dfrac{2}{3} u^{3/2} + c_1 \\\\ & = \dfrac{2}{3} \left(x^2 + x + 1\right)^{3/2} + c_1 \;\;\; \cdots (4a) \;\;\; \left[\text{From equation (5a)}\right] \end{aligned}$

$\begin{aligned} \text{Consider } I_2 & = \int \sqrt{x^2 + x + 1} \; dx \\\\ & = \int \sqrt{\left(x^2 + x + \dfrac{1}{4}\right) + 1 - \dfrac{1}{4}} \; dx \\\\ & = \int \sqrt{\left(x + \dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2} \; dx \\\\ & \left[\text{Note: } \int \sqrt{x^2 + a^2} = \dfrac{x}{2} \sqrt{x^2 + a^2} + \dfrac{a^2}{2} \log \left|x + \sqrt{x^2 + a^2}\right| + c \right] \\\\ & = \left(\dfrac{x + \dfrac{1}{2}}{2}\right) \sqrt{\left(x + \dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2} \\ & \hspace{6em} + \dfrac{\left(\sqrt{3} / 2\right)^2}{2} \log \left|x + \dfrac{1}{2} + \sqrt{\left(x + \dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2}\right| + c_2 \\\\ & = \left(\dfrac{2 x + 1}{4}\right) \sqrt{x^2 + x + 1} + \dfrac{3}{8} \log \left|x + \dfrac{1}{2} + \sqrt{x^2 + x + 1}\right| + c_2 \;\;\; \cdots (6) \end{aligned}$

In view of equations $(4a)$ and $(6)$, equation $(3)$ becomes

$I = \left(x^2 + x + 1\right)^{3/2} - \dfrac{7 \left(2 x + 1\right)}{8} \sqrt{x^2 + x + 1} - \dfrac{21}{16} \log \left|x + \dfrac{1}{2} + \sqrt{x^2 + x + 1}\right| + c$

where $c = \dfrac{3}{2} c_1 - \dfrac{7}{2} c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^2 \; e^x}{\left(x + 2\right)^2} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{x^2 \; e^x}{\left(x + 2\right)^2} \; dx \\\\ & = \int e^x \left\{\dfrac{x^2 + 4 x + 4 - 4 x - 4}{\left(x + 2\right)^2}\right\} \; dx \\\\ & = \int e^x \left\{\dfrac{\left(x + 2\right)^2 - 4 \left(x + 1\right)}{\left(x + 2\right)^2}\right\} \; dx \\\\ & = \int e^x \; dx - 4 \int e^x \left\{\dfrac{x + 1}{\left(x + 2\right)^2}\right\} \; dx \\\\ & = e^x - 4 \int e^x \left\{\dfrac{x + 2 - 1}{\left(x + 2\right)^2}\right\} \; dx \\\\ & = e^x - 4 \int e^x \left\{\dfrac{x + 2}{\left(x + 2\right)^2} - \dfrac{1}{\left(x + 2\right)^2}\right\} \; dx \\\\ & = e^x - 4 \int e^x \left\{\dfrac{1}{x + 2} - \dfrac{1}{\left(x + 2\right)^2}\right\} \; dx \\\\ & \left[\text{Note: } \int e^x \left[f \left(x\right) + f'\left(x\right) \; dx\right] = e^x f\left(x\right) + c\right] \\\\ & \text{Here, if } f\left(x\right) = \dfrac{1}{x + 2}, \text{ then } f'\left(x\right) = \dfrac{-1}{\left(x + 2\right)^2} \\\\ \therefore \; I & = e^x - \dfrac{4 \; e^x}{x + 2} + c \\\\ & = e^x \left(\dfrac{x + 2 - 4}{x + 2}\right) + c \\\\ & = e^x \left(\dfrac{x - 2}{x + 2}\right) + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int e^x \left(\dfrac{1 - \sin x}{1 - \cos x}\right) \; dx$

$\left[\begin{aligned} \text{Note: } & \sin x = 2 \sin \left(\dfrac{x}{2} \right) \cos \left(\dfrac{x}{2}\right) \\\\ & 1 - \cos x = 2 \sin^2 \left(\dfrac{x}{2}\right) \end{aligned}\right]$

$\begin{aligned} \text{Let } I & = \int e^x \left\{\dfrac{1 - \sin x}{1 - \cos x}\right\} \; dx \\\\ & = \int e^x \left\{\dfrac{1 - 2 \sin \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2}\right)}{2 \sin^2 \left(\dfrac{x}{2}\right)}\right\} \; dx \\\\ & = \int e^x \left\{\dfrac{1}{2 \sin^2 \left(\dfrac{x}{2}\right)} - \dfrac{\cos \left(\dfrac{x}{2}\right)}{\sin \left(\dfrac{x}{2}\right)}\right\} \; dx \\\\ & = \int e^x \left\{\dfrac{1}{2} \text{cosec}^2 \left(\dfrac{x}{2}\right) - \cot \left(\dfrac{x}{2}\right)\right\} \; dx \\\\ & \left[\text{Note: } \int e^x \left[f \left(x\right) + f'\left(x\right) \; dx\right] = e^x f\left(x\right) + c\right] \\\\ & \text{Here, if } f\left(x\right) = - \cot \left(\dfrac{x}{2}\right), \text{ then } f'\left(x\right) = \dfrac{1}{2} \text{cosec}^2 \left(\dfrac{x}{2}\right) \\\\ \therefore \; I & = \int e^x \left\{- \cot \left(\dfrac{x}{2} \right) + \dfrac{d}{dx} \left[- \cot \left(\dfrac{x}{2}\right)\right] \right\} \; dx \\\\ & = - e^x \cot \left(\dfrac{x}{2}\right) + c \end{aligned}$

Indefinite Integration

Evaluate $\int x^2 \sqrt{8 - x^6} \; dx$

Let $I = \int x^2 \sqrt{8 - x^6} \; dx$ $\;\;\; \cdots$ (1)

Let $x^3 = u$ $\;\;\; \cdots$ (2)

Differentiating equation (2) gives

$3 \; x^2 \; dx = du$ $\implies$ $x^2 \; dx = \dfrac{du}{3}$ $\;\;\; \cdots$ (2a)

Also, $x^6 = \left(x^3\right)^2 = u^2$ $\;\;\; \cdots$ (2b)

$\therefore$ In view of equations (2a) and (2b), equation (1) becomes

$\begin{aligned} I & = \dfrac{1}{3} \int \sqrt{8 - u^2} \; du \\\\ & = \dfrac{1}{3} \int \sqrt{\left(2 \sqrt{2}\right)^2 - u^2} \; du \\\\ & \left[\text{Note: } \int \sqrt{a^2 - x^2} \; dx = \dfrac{x}{2} \; \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right) + c \right] \\\\ & = \dfrac{1}{3} \left[\dfrac{u}{2} \; \sqrt{\left(2 \sqrt{2}\right)^2 - u^2} + \dfrac{\left(2 \sqrt{2}\right)^2}{2} \sin^{-1} \left(\dfrac{u}{2 \sqrt{2}}\right)\right] + c \\\\ & = \dfrac{1}{3} \left[\dfrac{x^3}{2} \; \sqrt{8 - x^6} + 4 \sin^{-1} \left(\dfrac{x^3}{2 \sqrt{2}}\right)\right] + c \;\;\; \left[\text{From equation (2)}\right] \end{aligned}$