Evaluate the following integrals:

Solution:

Assume $e^{x}-1=t^{2}$

$d\left(e^{x}-1\right)=d\left(t^{2}\right)$

$e^{x} \cdot d x=2 t \cdot d t$

$\Rightarrow \mathrm{dx}=\frac{2 t}{e^{\mathrm{x}}} \mathrm{dt}$

$e^{x}=t^{2}+1$

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

Substituting $t$ and $d t$

$\Rightarrow \int \sqrt{t^{2}} \cdot \frac{2 t}{t^{2}+1} d t$

$\Rightarrow \int t \cdot \frac{2 t}{t^{2}+1} d t$

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

$\Rightarrow 2 \int \frac{\mathrm{t}^{2}}{\mathrm{t}^{2}+1} \mathrm{dt}$

Add and subtract 1 in numerator

$\Rightarrow 2 \int \frac{t^{2}+1-1}{t^{2}+1} d t$

$\Rightarrow 2 \int \frac{t^{2}+1}{t^{2}+1} d t-2 \int \frac{1}{t^{2}+1} d t$

$\Rightarrow 2 \int d t-2 \int \frac{1}{t^{2}+1} d t$

$\Rightarrow \int \frac{1}{t^{2}+1} d t=\tan ^{-1} t+c$

$\Rightarrow 2 t-2 \tan ^{-1}(t)+c$

But $t=\left(e^{x}-1\right)^{1 / 2}$

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