Question

Evaluate the following integrals:

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2}{1-x^4}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{x^2}{1-x^4}=\frac{x^2}{\left(1-x^2\right)\left(1+x^2\right)} \\ =\frac{A}{1-x^2}+\frac{B}{1+x^2} \\ \Rightarrow x^2=A\left(1+x^2\right)+B\left(1-x^2\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 1=A-B\mathrm{and}0=A+B \\ \mathrm{or}A=\frac{1}{2}\mathrm{and}B=-\frac{1}{2} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{1}{2}}}{1-x^2}+\frac{{\displaystyle -\frac{1}{2}}}{1+x^2}\right)\mathrm{d}x \\ =\frac{1}{2}\int \frac{1}{1-x^2}\mathrm{d}x-\frac{1}{2}\int \frac{1}{1+x^2}\mathrm{d}x \\ =\frac{1}{2}\times \frac{1}{2}\log \left|\frac{1+x}{1-x}\right|-\frac{1}{2}{\tan }^{-1}x+c \\ =\frac{1}{4}\log \left|\frac{1+x}{1-x}\right|-\frac{1}{2}{\tan }^{-1}x+c \\ \mathrm{Hence},\int \frac{x^2}{1-x^4}\mathrm{d}x=\frac{1}{4}\log \left|\frac{1+x}{1-x}\right|-\frac{1}{2}{\tan }^{-1}x+c \end{gathered}$$