Evaluate the following integrals:

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

$d\left(\tan ^{-1} x\right)=d t$

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

Substituting $\mathrm{t}$ and $\mathrm{dt}$ in above equation we get

$\Rightarrow \int \frac{1}{\sqrt{t}} \mathrm{dt}$

$\Rightarrow \int \mathrm{t}^{-1 \backslash 2} \cdot \mathrm{dt}$

$\Rightarrow 2 \mathrm{t}^{1 \backslash 2}+\mathrm{c}$

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

$\Rightarrow 2\left(\tan ^{-1} x\right)^{1 / 2}+c$