Question

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

Answer 1

We have,

$I=\int \frac{x^2}{x^4+x^2+1}dx$

$I=\int \frac{x^2}{(x^2-x+1)(x^2+x+1)}dx$

$I=\frac{1}{2}\int \frac{x}{(x^2-x+1)}dx-\frac{1}{2}\int \frac{x}{(x^2+x+1)}dx$

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

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

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

$I=\frac{1}{4}\ln (x^2-x+1)+\frac{1}{4}\left[\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]-\frac{1}{4}\ln (x^2+x+1)+\frac{1}{4}\left[\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]+C$

$I=\frac{1}{4}\ln (x^2-x+1)+\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)-\frac{1}{4}\ln (x^2+x+1)+\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$

$I=\frac{1}{4}\ln (x^2-x+1)-\frac{1}{4}\ln (x^2+x+1)+\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$

Hence, this is the answer.