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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 19

$\log \left|\frac{\left(x^{2}-1\right)^{3}}{x}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{5 x^{2}-1}{x(x-1)(x+1)} d x$

Explanation:

Let

$I=\int \frac{5 x^{2}-1}{x(x-1)(x+1)} d x$
\begin{aligned} &\frac{5 x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} \\ &5 x^{2}+1=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1) \\ &\text { At } x=1 \\ &5(1)+1=A(0)+B(1)(2)+C(0) \\ &6=2 B \\ &B=3 \end{aligned}
\begin{aligned} &\text { At } x=-1 \\ &5(1)+1=A(0)+B(0)+C(-1)(-2) \\ &6=2 C \\ &C=3 \end{aligned}
\begin{aligned} &\text { At } x=0 \\ &5(0)+1=A(-1)(1)+0+0 \\ &1=-A \\ &A=-1 \end{aligned}
\begin{aligned} &\frac{5 x^{2}-1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{3}{x-1}+\frac{3}{x+1} \\ &I=-\int \frac{1}{x} d x+3 \int \frac{1}{x-1} d x+3 \int \frac{1}{x+1} d x \\ &=-\log |x|+3 \log |x-1|+3 \log |x+1|+C \end{aligned}

Using the formulas,$(x+y)(x-y)=x^{2}-y^{2} \& \log x+\log y=\log\left ( xy \right )$
$I=\log \left|\frac{\left(x^{2}-1\right)^{3}}{x}\right|+C$