Question

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

Answer 1

$\frac{e^{2x}-1}{e^{2x}+1}$
Dividing numerator and denominator by $e^x$, we obtain
$\frac{\frac{(e^{2x}-1)}{e^x}}{\frac{(e^{2x}+1)}{e^x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
Put $e^x+e^{-x}=t$
$\therefore (e^x-e^{-x})dx=dt$
$\Rightarrow \int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$
$=\int \frac{dt}{t}$
$=\log |t|+C=\log |e^x+e^{-x}|+C$