#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 8

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}+1}{x\left(x^{2}-1\right)} d x \\ &I=\int \frac{\left(x^{2}-1\right)+2}{x\left(x^{2}-1\right)} d x \end{aligned}                 [Add and subtract 1]

\begin{aligned} &I=\int\left(\frac{1}{x}+\frac{2}{x\left(x^{2}-1\right)}\right) d x \\ &I=\int \frac{1}{x} d x+2 \int \frac{1}{x(x-1)(x+1)} d x \end{aligned}…(applying the formula $a^{2}-b^{2}$)

$I=\log |x|+2 I_{1}$            (1)

Where

\begin{aligned} &I_{1}=\int \frac{1}{x(x-1)(x+1)} d x \\ &\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} \\ &\frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)} \\ &1=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1) \end{aligned}

\begin{aligned} &\text { At } x=0 \\ &1=A(-1)(1)+0+0 \\ &A=-1 \end{aligned}

\begin{aligned} &\text { At } x=1 \\ &1=0+B(1)(2)+0 \\ &1=2 B \\ &B=\frac{1}{2} \end{aligned}

\begin{aligned} &\text { At } x=-1 \\ &1=0+0+C(-1)(-2) \\ &1=2 C \\ &C=\frac{1}{2} \end{aligned}

\begin{aligned} &\frac{1}{(x-1)(x+1) x}=\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)} \\ &I_{1}=\int\left[\frac{-1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right] d x \\ &I_{1}=-\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{1}{x+1} d x \\ &I_{1}=-\log |x|+\frac{1}{2} \log |x-1|+\frac{1}{2} \log |x+1|+C \end{aligned}

Equation (1)

$I=\log |x|-2 \log |x|+\log |x-1|+\log |x+1|+C$

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