Evaluate the following integrals:

Let $I=\int \frac{x}{x^{4}+2 x^{2}+3} d x$

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

$\Rightarrow 2 x d x=d t$

$I=\frac{1}{2} \int \frac{1}{t^{2}+2 t+3} d t$

$=\frac{1}{2} \int \frac{1}{t^{2}+2 t+1-1+3} d t$

$=\frac{1}{2} \int \frac{1}{(\mathrm{t}+1)^{2}+2} \mathrm{dt}$

Put $t+1=u$               .......(ii)

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

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

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

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

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

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