Question
Evaluate the following integrals:
$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$

Answer

$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Let $\text{e}^\text{x}-1=\text{t}^2$
$\Rightarrow\text{e}^\text{x}=\text{t}^2+1$
$\text{e}^\text{x}=2\text{t}\frac{\text{dt}}{\text{dx}}$
$\text{dx}=\frac{2\text{t dt}}{\text{e}^\text{x}}$
$\text{dx}=\frac{2\text{t dt}}{\text{t}^2+1}$
Now, $\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
$=\int\frac{\text{t. 2t dt}}{\text{t}^2+1}$
$=2\int\frac{\text{t}^2\text{ dt}}{\text{t}^2+1}$
$=2\int\Big(\frac{\text{t}^2+1-1}{\text{t}^2+1}\Big)\text{ dt}$
$=2\int\text{dt}-2\int\frac{\text{dt}}{\text{t}^2+1}$
$=2\text{t}-2\tan^{-1}(\text{t})+\text{C}$
$=2\sqrt{\text{e}^\text{x}-1}-2\tan^{-1}\big(\sqrt{\text{e}^\text{x}-1}\big)+\text{C}$

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