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 $c = c_1 - c_2 + c_3$