Question

Solve $\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx$

Answer 1

Given $\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx$

Suppose $u=e^{2x}+e^{-2x}$ $du=2(e^{2x}-e^{-2x})dx$

$=\int \frac{1}{2u}du$

$=\frac{1}{2}\int \frac{1}{u}du$

$=\frac{\ln \left(u\right)}{2}$

$=\frac{\ln (e^{2x}+e^{-2x})}{2}$

$=\frac{2x+(-2x)}{2}$

$=\frac{2x-2x}{2}$

$=0$