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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 16

$\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x}{\left(x^{2}+1\right)(x+1)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{x}{\left(x^{2}+1\right)(x+1)} d x \\ &\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+1} \end{aligned}

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

Comparing the coefficient

$A+B=0$           (1)

$A=-B$

$C-B=1$           (2)

$A-C=0$          (3)

$A=C$

Equation (2)

\begin{aligned} &A+A=1 \\ &2 A=1 \\ &A=\frac{1}{2} \\ &B=\frac{-1}{2} \\ &C=\frac{1}{2} \end{aligned}

\begin{aligned} &\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{1}{2(x-1)}+\frac{\frac{-1}{2} x+\frac{1}{2}}{\left(x^{2}+1\right)} \\ &=\frac{1}{2(x-1)}+\frac{1-x}{2\left(x^{2}+1\right)} \end{aligned}

\begin{aligned} &I=\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1-x}{x^{2}+1} d x \\ &I=\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x \\ &I=\frac{1}{2} \log |x-1|+\frac{1}{2} \tan ^{-1} x-\frac{1}{4} \log \left|x^{2}+1\right|+C \end{aligned}