Question

The value of $\int \sqrt{\frac{ e ^{ x }-1}{ e ^{ x }+1}} dx$ is equal to

• A. $ \ln \left(e^x+\sqrt{e^{2 x}-1}\right)-\sec ^{-1}\left(e^x\right)+C$
• B. $\ln \left(e^x+\sqrt{e^{2 x}-1}\right)+\sec ^{-1}\left(e^x\right)+C$
• C. $\ln \left(e^x-\sqrt{e^{2 x}-1}\right)-\sec ^{-1}\left(e^x\right)+C$
• D. None of these

Answer 1

Answer: (A)

$I=\int \frac{e^x-1}{\sqrt{e^{2 x}-1}} d x$
$= \int \frac{e^x}{\sqrt{e^{2 x}-1}} d x-\int \frac{e^x}{e^x \sqrt{e^{2 x}-1}} d x$
put $e^x=t \Rightarrow e^x d x=d t$
$\Rightarrow I=\int \frac{d t}{\sqrt{t^2-1}}-\int \frac{d t}{t \sqrt{t^2-1}}$
$=e n\left|t+\sqrt{t^2-1}\right|-\sec ^{-1}(t)+C$,
where $t=e^x$