Question

Solve:-
$\int \frac{dx}{\sqrt{e^{2x}-1}}$

Answer 1

$I=\int \frac{dx}{\sqrt{(e^x{)}^2}-1}$

Let $\sqrt{e^{2x}-1}=t=\Rightarrow e^{2x}=t^2+1$

$\frac{\left(2e^{2x}\right)}{2\sqrt{e^{2x}-1}}=\frac{dt}{dx}\Rightarrow \frac{dx}{\sqrt{e^{2x}-1}}=\frac{dt}{t^2+1}$

$=\int \frac{dt}{t^2+1}={\tan }^{-1}\left(t\right)+c$

Let the volume of $t$

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