Evaluate the following integrals:

Let $I=\int \frac{x}{x^{4}-x^{2}+1} d x$

Let $\mathrm{x}^{2}=\mathrm{t} \ldots \ldots \cdots$ (i)

$\Rightarrow 2 x d x=d t$

$I=\frac{1}{2} \int \frac{1}{t^{2}-t+1} d t$

$=\frac{1}{2} \int \frac{1}{t^{2}-2 t\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d t$

$=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}} d t$

Put $\mathrm{t}-1 / 2=\mathrm{u}$             ............(ii)

$\Rightarrow \mathrm{dt}=\mathrm{du}$

$I=\frac{1}{2} \int \frac{1}{(u)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d u$

$I=\frac{1}{2 \frac{\sqrt{3}}{2}} \tan ^{-1} \frac{u}{\frac{\sqrt{3}}{2}}+c$

$\left[\right.$ since, $\left.\int \frac{1}{x^{2}+(a)^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right]$

$I=\frac{1}{2 \frac{\sqrt{3}}{2}} \tan ^{-1} \frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}+c$ [using (i)]

$I=\frac{1}{\sqrt{3}} \tan ^{-1} \frac{2 x^{2}-1}{\sqrt{3}}+c$ [using (ii)]