Question

Evaluate the following integrals:

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2}{x^4+x^2-2}\mathrm{d}x \\ \mathrm{We}\mathrm{express} \\ \frac{x^2}{x^4+x^2-2}=\frac{x^2}{x^4+2x^2-x^2-2} \\ =\frac{x^2}{\left(x^2+2\right)\left(x^2-1\right)} \\ =\frac{A}{x^2+2}+\frac{B}{x^2-1} \\ \Rightarrow x^2=A\left(x^2-1\right)+B\left(x^2+2\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 1=A+B\mathrm{and}0=-A+2B \\ \mathrm{or}A=\frac{2}{3}\mathrm{and}B=\frac{1}{3} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{2}{3}}}{x^2+2}+\frac{{\displaystyle \frac{1}{3}}}{x^2-1}\right)\mathrm{d}x \\ =\frac{2}{3}\int \frac{1}{x^2+2}\mathrm{d}x+\frac{1}{3}\int \frac{1}{x^2-1}\mathrm{d}x \\ =\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{x}{\sqrt{2}}+\frac{1}{6}\log \left|\frac{x-1}{x+1}\right|+c \\ \mathrm{Hence},\int \frac{x^2}{x^4+x^2-2}\mathrm{d}x=\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{x}{\sqrt{2}}+\frac{1}{6}\log \left|\frac{x-1}{x+1}\right|+c \end{gathered}$$