Evaluate the following integrals:

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

$\tan ^{-1} \mathrm{x}=\mathrm{t} ; \mathrm{x}=\operatorname{tant} \int \frac{\mathrm{x}^{2} \tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}$

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

$I=\int t \tan ^{2} t d t$

We know that, $\tan ^{2} \mathrm{t}=\sec ^{2} \mathrm{t}-1$

$=\int \mathrm{t}\left(\sec ^{2} \mathrm{t}-1\right) \mathrm{dt}$

$=\int \mathrm{tsec}^{2} \mathrm{t} \mathrm{dt}-\int \mathrm{tdt}$

Using integration by parts,

$=\left(\mathrm{t} \int \sec ^{2} \mathrm{tdt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \sec ^{2} \mathrm{tdt}\right)-\frac{\mathrm{t}^{2}}{2}$

$=\left(\mathrm{ttan} \mathrm{t}-\int \tan \mathrm{t} \mathrm{dt}\right)-\frac{\mathrm{t}^{2}}{2}$

$=(\mathrm{t} \tan \mathrm{t}-\log |\sec \mathrm{t}|)-\frac{\mathrm{t}^{2}}{2}+\mathrm{c}$

$=\left[x \tan ^{-1} x+\log \left|\sqrt{1+x^{2}}\right|\right]-\frac{\tan ^{2} x}{2}+c$

$=x \tan ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|-\frac{\tan ^{2} x}{2}+c$