#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 57

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

Equating both side

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

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

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

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