Prove

$\frac{e^{2 x}-1}{e^{2 x}+1}$

Dividing numerator and denominator by ex, we obtain

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

Let $e^{x}+e^{-x}=t$

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

$\Rightarrow \int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$

$=\int \frac{d t}{t}$

$=\log |t|+\mathrm{C}$

$=\log \left|e^{x}+e^{-x}\right|+\mathrm{C}$