#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 67

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

\begin{aligned} &\text { Let }\\ &I=\int \frac{x^{2}}{1-x^{4}} d x \end{aligned}

\begin{aligned} &\text { Let } x^{2}=y \\ &\frac{y}{1-y^{2}}=\frac{y}{(1+y)(1-y)}=\frac{A}{1+y}+\frac{B}{1-y} \\ &y=A(1-y)+B(1+y) \end{aligned}
$\begin{gathered} \text { At } y=1 \\ 1=2 B \\ B=\frac{1}{2} \end{gathered}$
\begin{aligned} &\text { At } y=-1 \\ &-1=2 A \\ &A=\frac{-1}{2} \\ &I=\frac{-1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{1-x^{2}} d x \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{2} \int \frac{1}{1-x^{2}} d x \end{aligned}
\begin{aligned} &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{2} \int \frac{1}{(1+x)(1-x)} d x \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{2} \cdot \frac{1}{2} \int \frac{2+x-x}{(1+x)(1-x)} d x \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \int \frac{(1+x)+(1-x)}{(1+x)(1-x)} d x \end{aligned}
\begin{aligned} &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \int \frac{1}{1-x} d x+\frac{1}{4} \int \frac{1}{1+x} d x \\ &I=\frac{-1}{2} \tan ^{-1} x-\frac{1}{4} \log (1-x)+\frac{1}{4} \log (1+x)+C \\ &I=\frac{-1}{2} \tan ^{-1} x+\frac{1}{4} \log \left(\frac{1+x}{1-x}\right)+C \end{aligned}