Question

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

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

Answer 1

Answer: (A)

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