#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 37

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

Equating similar terms

\begin{aligned} &A+B=0 \\ &B=-A \end{aligned}                       (1)

\begin{aligned} &A+C=0 \\ &C=-A \end{aligned}                       (2)

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

Equation (1)

$B=\frac{1}{2}$

Equation (2)

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