Question

Evaluate: $\int \frac{x^2}{x^4+x^2-2}dx$.

Answer 1

$I=\int \frac{x^2}{x^4+x^2-2}dx$
$I=\int \frac{x^2}{x^4+{2x}^2-x^2-2}dx$
$I=\int \frac{x^2}{x^2(x^2+2)-1(x^2+2)}dx$
$I=\int \frac{x^2}{(x^2+2)(x^2-1)}dx$
$\frac{x^2}{(x^2+2)(x^2-1)}=\frac{A}{x^2+2}+\frac{B}{x^2-1}$
$\Rightarrow x^2=A(x^2-1)+B(x^2+2)$
Put $x=1$
$1=0+3B⟹B=\frac{1}{3}$
Put $x=0\Rightarrow 0=-A+2B\Rightarrow A=2B\Rrightarrow A=\frac{2}{3}$
Therefore, $I=\frac{2}{3}\int \frac{1}{x^2+2}dx+\frac{1}{3}\int \frac{dx}{x^2-1}$
$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}$
$\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln \left[\frac{x-a}{x+a}\right]$
Thus $I=\frac{2}{3}\times \frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}+\frac{1}{3}\times \frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C$
$I=\frac{\sqrt{2}}{3}{\tan }^{-1}\frac{x}{\sqrt{2}}+\frac{1}{6}\ln \left|\frac{x-1}{x+1}\right|+C$