Question

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

Answer 1

Given : $\int \frac{e^{2x}-1}{e^{2x}+1}dx$
Divide numerator and denominator by $e^x,$ we get
$=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$
Let $e^x+e^{-x}=t$
Differentiating both sides w.r.t.$x$
$e^x-e^{-x}=\frac{dt}{dx}\Rightarrow (e^x-e^{-x})dx=dt$
On substituting
$=\int \frac{1}{t}dt$
$=\log |t|+C$
$=\log |e^x+e^{-x}|+C$
$=\log (e^x+e^{-x})+C$
$(\because e^x+e^{-x}>0)$
Where $C$ is constant of integration