Definite Integration

Obtain $\displaystyle \int \limits_{\log_{a}2}^{\log_{a}4} a^x \; dx$ as the limit of a sum.

Let $f\left(x\right) = a^x$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[\log_{a} 2, \log_{a} 4\right]$.

Lower limit $= a = \log_{a} 2$; Upper limit $= b = \log_{a} 4$

Divide $\left[\log_{a} 2, \log_{a} 4\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{\log_{a} 4 - \log_{a} 2}{n} = \dfrac{\log_{a} 2}{n}$

$\implies$ $n h = \log_{a} 2$ $\;\;\; \cdots \; (2)$

Now, $f\left(a + kh\right) = f\left(\left(\log_{a} 2\right) + kh\right) = a^{\left(\log_{a} 2\right) + k h} = a^{\log_{a} 2} \times a^{k h} = 2 a^{k h}$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]

Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$

By definition, $\displaystyle \int \limits_{\log_{a} 2}^{\log_{a} 4} a^{x} \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{\log_{a} 2}^{\log_{a} 4} a^x \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} 2 a^{k h}\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\ & = \lim\limits_{h \to 0} 2 h \left[a^{h} + a^{2 h} + a^{3 h} + \cdots + a^{n h}\right] \\\\ & \left[\begin{aligned} \text{Note: } & a^{h} + a^{2 h} + a^{3 h} + \cdots + a^{n h} \text{ is a geometric series} \\\\ & \text{with first term} = a^{h} \text{ and common ratio } = a^{h} \\\\ & \text{Sum to n terms of a geometric series with first term A} \\\\ & \text{ and common ratio R } = \dfrac{A \left(R^n - 1\right)}{R - 1} \text{ when } R > 1 \end{aligned}\right] \\\\ & = \lim\limits_{h \to 0} 2 h \left[\dfrac{a^{h} \left(a^{n h} - 1\right)}{a^{h} - 1}\right] \\\\ & = \lim\limits_{h \to 0} 2 \; a^{h} \left[\dfrac{a^{\log_{a} 2} - 1}{\dfrac{a^{h} - 1}{h}}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \dfrac{2 \times a^{0} \times \left(2 - 1\right)}{\log_{e} a} \\\\ & = 2 \times 1 \times 1 \times \log_{a} e \\\\ & = 2 \log_{a} e \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{0}^{1} e^{2 - 3 x} \; dx$ as the limit of a sum.

Let $f\left(x\right) = e^{2 - 3x}$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[0,1\right]$.

Lower limit $= a = 0$; Upper limit $= b = 1$

Divide $\left[0,1\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{1 - 0}{n} = \dfrac{1}{n}$

$\implies$ $n h = 1$ $\;\;\; \cdots \; (2)$

Now, $f\left(a + kh\right) = f\left(kh\right) = e^{2 - 3 k h} = \dfrac{e^2}{e^{3 k h}}$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]

Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$

By definition, $\displaystyle \int \limits_{0}^{1} e^{2 - 3x} \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{0}^{1} e^{2 - 3x} \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} \dfrac{e^2}{e^{3kh}}\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\ & = \lim\limits_{h \to 0} e^2 h \left[\dfrac{1}{e^{3h}} + \dfrac{1}{e^{6h}} + \dfrac{1}{e^{9h}} + \cdots + \dfrac{1}{e^{nh}}\right] \\\\ & \left[\begin{aligned} \text{Note: } & \dfrac{1}{e^{3h}} + \dfrac{1}{e^{6h}} + \dfrac{1}{e^{9h}} + \cdots + \dfrac{1}{e^{nh}} \text{ is a geometric series} \\\\ & \text{with first term} = \dfrac{1}{e^{3h}} \text{ and common ratio } = \dfrac{1}{e^{3h}} \\\\ & \text{Sum to n terms of a geometric series with first term A} \\\\ & \text{ and common ratio R } = \dfrac{A \left(1 - R^n\right)}{1 - R} \text{ when } R < 1 \end{aligned}\right] \\\\ & = \lim\limits_{h \to 0} e^2 h \left[\dfrac{\dfrac{1}{e^{3h}} \left(1 - \dfrac{1}{e^{3nh}}\right)}{1 - \dfrac{1}{e^{3h}}}\right] \\\\ & = \lim\limits_{h \to 0} e^2 h \left[\dfrac{1}{e^{3h}} \times \dfrac{\left(1 - \dfrac{1}{e^3}\right)}{e^{3h} - 1} \times e^{3h}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \dfrac{e^3 -1}{e} \; \lim\limits_{h \to 0} \dfrac{1}{\left(\dfrac{e^{3h} - 1}{3h}\right) \times 3} \\\\ & = \dfrac{1}{3} \left(\dfrac{e^3 - 1}{e}\right) \times \dfrac{1}{\log_{e} e} \\\\ & = \dfrac{1}{3} \left(\dfrac{e^3 - 1}{e}\right) \end{aligned}$

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