#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 65

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

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{x^{4}}{(x-1)\left(x^{2}+1\right)} d x$

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{4}}{(x-1)\left(x^{2}+1\right)} d x \\ &I=\int \frac{x^{4}}{x^{3}-x^{2}+x-1} d x \\ &I=\frac{x\left(x^{3}-x^{2}+x-1\right)+1\left(x^{3}-x^{2}+x-1\right)+1}{\left(x^{3}-x^{2}+x-1\right)} \\ &I=x+1+\frac{1}{(x-1)\left(x^{2}+1\right)} \end{aligned}

Let,

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

Put $x=1$

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

Put $x= 0$

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

Put $x= -1$

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

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