Question

$\int \frac{e^{2x}-1}{e^{2x}+1}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{1}{2}\ln |e^{2x}-e^{-2x}|+C$
• B. $\ln |e^x+e^{-x}|+C$
• C. $\ln |e^{2x}+e^{-2x}|+C$
• D. $\ln |e^x-e^{-x}|+C$

Answer 1

The correct option is B $\ln |e^x+e^{-x}|+C$

$I=\int \frac{e^{2x}-1}{e^{2x}+1}dx$
$[$Dividing numerator and denominator by $e^x]$
$\Rightarrow I=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$
Putting $e^x+e^{-x}=t$
$\Rightarrow (e^x-e^{-x})dx=dt\therefore I=\int \frac{1}{t}dt=\ln |t|+C\Rightarrow I=\ln |e^x+e^{-x}|+C$