#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 126 Maths Textbook Solution.

Answer:$2 \log |x|-\frac{1}{4} \log |x+1|-\frac{1}{4} \log |x-1|-\frac{3}{4} \log \left|x^{2}+1\right|+c$

Given: $\int \frac{x^{2}-2}{x^{5}-x} d x$

Hint: using partial fraction

Explanation let $I=\int \frac{x^{2}-2}{x^{5}-x} d x$

\begin{aligned} &=\int \frac{x^{2}-2}{x\left(x^{4}-1\right)} d x=\int \frac{x^{2}-2}{x\left(x^{2}-1\right)\left(x^{2}+1\right)} \\ &I=\int \frac{x^{2}-2}{x(x+1)(x-1)\left(x^{2}+1\right)} d x \\ &\frac{x^{2}-2}{x(x+1)(x-1)\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}+\frac{D x+E}{x^{2}+1} \end{aligned}

Multiply by $x(x+1)(x-1)\left(x^{2}+1\right)$

$\inline x^{2}-2=A(x+1)(x-1)\left(x^{2}+1\right)+B(u)(u-1)\left(x^{2}+1\right)+C(x)(x+1)\left(x^{2}+1\right)+(D x+E)(x)(x+1)(x-1)$

\inline \begin{aligned} &\text { putting } x=0 \\ &0-2=A(1)(-1)(1)+B(0)+C(0)+(D x+6)(10) \\ &-2=-A \Rightarrow A=2 \end{aligned}

\inline \begin{aligned} &\text { putting } x=1 \\ &1-2=A(2)(0)(2)+B(0)+C(1)(2)(2)+(D x+6)(0) \\ &-1=4 C=>C=\frac{-1}{4} \end{aligned}

\inline \begin{aligned} &\text { putting } x=-1\\ &1-2=A(0)+B(-1)(-2)(2)+C(0)+D(0)\\ &-1=>4 B \Rightarrow B=\frac{-1}{4} \end{aligned}

$\inline \text { On solving }, D=-\frac{3}{2} \& E=0$

$\inline \int \frac{x^{2}-2}{x^{5}-x} d x=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1} d x-\frac{3}{2} \int \frac{x}{x^{2}+1} d x$

$\int \frac{x^{2}-2}{x^{5}-x} d x=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1} d x-\frac{3}{2} \int \frac{x}{x^{2}+1} d x$

$=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1}-\frac{3}{4} \int \frac{2 x d x}{x^{2}+1}$

$=2 \log |x|-\frac{1}{4} \log |x+1|-\frac{1}{4} \log |x-1|-\frac{3}{4} \log \left|x^{2}+1\right|+c$