# Evaluate the following integral:

Question:

Evaluate the following integral:

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

Solution:

Let, $\mathrm{I}=\int \frac{\mathrm{x}^{2}}{1-\mathrm{x}^{4}} \mathrm{dx}$

Let, $\frac{\mathrm{x}^{2}}{1-\mathrm{x}^{4}}=\frac{\mathrm{A}}{1-\mathrm{x}}+\frac{\mathrm{B}}{1+\mathrm{x}}+\frac{\mathrm{C}}{1+\mathrm{x}^{2}}$

$\Rightarrow x^{2}=A(1+x)\left(x^{2}+1\right)+B(1-x)\left(x^{2}+1\right)+c(x+1)(1-x)$

For, $x=1, A=\frac{1}{4}$

For, $\mathrm{x}=-1, \mathrm{~B}=\frac{1}{4}$

For, $x=0, C=-\frac{1}{2}$

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

$\Rightarrow I=-\frac{1}{4} \log |1-x|+\frac{1}{4} \log |1+x|-\frac{1}{2} \tan ^{-1} x+c$

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

Hence, $\int \frac{\mathrm{x}^{2}}{1-\mathrm{x}^{4}} \mathrm{dx}=\frac{1}{4} \log \left|\frac{1+\mathrm{x}}{1-\mathrm{x}}\right|-\frac{1}{2} \tan ^{-1} \mathrm{x}+\mathrm{c}$

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