Question

Integrate: $\int \frac{x}{x^4-x^2+1}dx$

Answer 1

Solution:-

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

Let $x^2=t\Rightarrow xdx=\frac{dt}{2}$

$\therefore \int \frac{x}{x^4-x^2+1}dx=\frac{1}{2}\int \frac{dt}{t^2-t+1}$

$\Rightarrow =\frac{1}{2}\int \frac{dt}{{(t-\frac{1}{2})}^2-\frac{1}{4}+1}$

$\Rightarrow =\frac{1}{2}\int \frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$\Rightarrow =\frac{1}{2}\left[\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}{\tan }^{-1}\frac{(t-\frac{1}{2})}{\left(\frac{\sqrt{3}}{2}\right)}\right]+C$

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

As $t=x^2$

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