Question

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

Answer 1

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

For Partial fraction, Put $x^2=t$

$\Rightarrow \frac{t}{t^2+t-2}=\frac{t}{(t+2)(t-1)}$

Let $\frac{t}{(t+2)(t-1)}=\frac{A}{t+2}+\frac{B}{t-1}$ ... (i)

$\Rightarrow t=A(t-1)+B(t+2)$

Put $t=-2$

$\Rightarrow -2=A(-2-1)$

$\Rightarrow -3A=-2\Rightarrow A=\frac{2}{3}$

Put $t=1$

$\Rightarrow 1=B(1+2)$

$\Rightarrow 3B=1$

$\Rightarrow B=\frac{1}{3}$

From (i), we have

$\frac{t}{(t+2)(t-1)}=\frac{\frac{2}{3}}{t+2}+\frac{\frac{1}{3}}{t-1}$

$\therefore \int \frac{x^2}{x^4+x^2-2}dx=\frac{2}{3}\int \frac{1}{x^2+(\sqrt{2}{)}^2}dx+\frac{1}{3}\int \frac{1}{x^2-1^2}dx$

$=\frac{2}{3}\times \frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{3}\times \frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+c$

$=\frac{\sqrt{2}}{3}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{6}\log \left|\frac{x-1}{x+1}\right|+c$