Question

Integrate the function $\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

Answer 1

Let $e^{2x}+e^{-2x}=t$
$\therefore (2e^{2x}-2e^{-2x})dx=dt$
$\Rightarrow 2(e^{2x}-e^{-2x})dx=dt$
$\Rightarrow \int \left(\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\right)dx=\int \frac{dt}{2t}$
$=\frac{1}{2}\int \frac{1}{t}dt$
$=\frac{1}{2}\log |t|+C$
$=\frac{1}{2}\log |e^{2x}+e^{-2x}|+C$