Evaluate the following integrals:

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

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

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

$\Rightarrow \frac{2+\mathrm{x}^{2}}{\mathrm{x}^{2}+1}=\mathrm{dt}$

Substituting $\mathrm{t}$ and $\mathrm{dt}$

$\Rightarrow \int 5^{t} d t$

$\Rightarrow \frac{5^{t}}{\log 5}+c$

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

$\Rightarrow \frac{5^{x+\tan ^{-1} x}}{\log 5}+c$