Evaluate the following integrals:

Let $I=\int \frac{x \tan ^{-1} x}{\left(1+x^{2}\right)^{\frac{3}{2}}} d x$

$\tan ^{-1} \mathrm{x}=\mathrm{t}$

$\frac{1}{1+x^{2}} d x=d t$

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

We know that, $\sqrt{1+\tan ^{2} t}=\sec t$

$=\int \frac{\mathrm{t} \tan t}{\sec t} \mathrm{dt}$

$=\int \mathrm{t} \frac{\sin t}{\cos t} \cos t \mathrm{dt}$

$=\int \mathrm{t} \sin \mathrm{t} \mathrm{dt}$

Using integration by parts,

$=\mathrm{t} \int \sin \mathrm{t} \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \sin \mathrm{t} \mathrm{dt}$

$=-t \cos t+\int \cos t d t$

$=-t \cos t+\sin t+c$

Substitute value for $\mathrm{t}$

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