Evaluate the following integrals:

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

$\mathrm{d} \mathrm{x}=\sec ^{2} \mathrm{t} \mathrm{dt}$

$d x=\sec ^{2} t d t$

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

We know that $\frac{1-\tan ^{2} t}{1+\tan ^{2} t}=\cos 2 t$

$=\int \cos ^{-1}(\cos 2 t) \sec ^{2} t d t$

$=\int 2 t \sec ^{2} t d t$

Using integration by parts,

$=2\left[\mathrm{t} \int \sec ^{2} \mathrm{t} \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \mathrm{sec}^{2} \mathrm{t} \mathrm{dt}\right]$

$=2\left[\mathrm{t} \tan \mathrm{t}-\int \tan \mathrm{t} \mathrm{dt}\right]$

$=2[\mathrm{t} \tan \mathrm{t}-\log \sec \mathrm{t}]+\mathrm{c}$

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

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