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,
$=\left(\mathrm{t} \int \sec ^{2} \mathrm{tdt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \sec ^{2} \mathrm{tdt}\right)-\frac{\mathrm{t}^{2}}{2}$
$=\left(\mathrm{ttan} \mathrm{t}-\int \tan \mathrm{t} \mathrm{dt}\right)-\frac{\mathrm{t}^{2}}{2}$
$=(\mathrm{t} \tan \mathrm{t}-\log |\sec \mathrm{t}|)-\frac{\mathrm{t}^{2}}{2}+\mathrm{c}$
$=\left[x \tan ^{-1} x+\log \left|\sqrt{1+x^{2}}\right|\right]-\frac{\tan ^{2} x}{2}+c$
$=x \tan ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|-\frac{\tan ^{2} x}{2}+c$
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