Question

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

Answer 1

We have,

$\begin{array}{c}\int \frac{x^2}{x^4+x^2+1}dx\\ =\frac{1}{2}\int \frac{\left(x^2+1\right)+\left(x^2-1\right)}{x^4+x^2+1}dx\\ =\frac{1}{2}\int \frac{1+x^2}{x^4+x^2+1}dx+\frac{1}{2}\int \frac{x^2-1}{x^4+x^2+1}dx\\ =\frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)dx}{{\left(x-\frac{1}{x}\right)}^2+3}+\frac{1}{2}\int \frac{\left(1-\frac{1}{x^2}\right)dx}{{\left(x+\frac{1}{x}\right)}^2-1}\\ =\frac{1}{2}\int \frac{dt}{t^2+3}+\frac{1}{2}\int \frac{dt}{v^2-1}\\ t=x-\frac{1}{x}v=x+\frac{1}{x}\\ =\frac{1}{2}\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t}{\sqrt{3}}\right)+\frac{1}{2}.\frac{1}{2}\ln \left|\frac{v-1}{v+1}\right|+C\\ =\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+\frac{1}{4}\ln \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+C\end{array}$

Hence, this is the answer.