Question

$\int \frac{x}{x^4+x^2+1}dx$, Integration gives
$\frac{1}{\sqrt{\left(k\right)}}{\tan }^{-1}\frac{2x^2+1}{\sqrt{\left(k\right)}}$
find $k^2.$

Answer 1

Substitute $x^2=t$

$\therefore 2xdx=dt$

$\therefore I=\frac{1}{2}\int \frac{dt}{t^2+t+1}$ $=\frac{1}{2}\int \frac{dt}{t^2+t+\frac{1}{4}+(1-\frac{1}{4})}=\frac{1}{2}\int \frac{dt}{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$ $=\frac{1}{2.\frac{\sqrt{\left(3\right)}}{2}}\tan -1\frac{t+\frac{1}{2}}{\sqrt{\left(3\right)/2}}=\frac{1}{\sqrt{\left(3\right)}}{\tan }^{-1}\frac{2x^2+1}{\sqrt{\left(3\right)}}.$