Question

Integrate the following function $w.r.tx$
$\frac{x}{x^4+x^2+1}$

Answer 1

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

Let $x^2=u\Rightarrow 2xdx=du$

or, $xdx=\frac{du}{2}$

$\therefore I=\frac{1}{2}\int \frac{du}{u^2+u+1}$

$=\frac{1}{2}\int \frac{du}{u^2+2\times \frac{1}{2}\times u+\frac{1}{4}+1-\frac{1}{4}}$

$=\frac{1}{2}\int \frac{du}{{(u+\frac{1}{2})}^2}+\frac{3}{4}$

$=\frac{1}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left|\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|+C$, ($c=$ constant of integration)

$=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2u+1}{\sqrt{3}}\right)+C$

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