Question

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

• A. $ln(e^x+\sqrt{e^{2x}-1}-se{c}^{-1}(e^x)+C$
• B. $ln(e^x+\sqrt{e^{2x}-1}+si{n}^{-1}(e^x)+C$
• C. $ln(e^x-\sqrt{e^{2x}-1}-se{c}^{-1}(e^x)+C$
• D. none of these

Answer 1

The correct option is B $ln(e^x+\sqrt{e^{2x}-1}+si{n}^{-1}(e^x)+C$

$\int \sqrt{\frac{e^x-1}{e^x+1}}dx=\int \sqrt{\frac{e^x-1}{e^x+1}.\frac{e^x-1}{e^x-1}}dx=\int \frac{e^x-1}{\sqrt{e^{2x}-1}}dx$

$=\int [\frac{e^x.}{\sqrt{(e^x{)}^2-1}}+\frac{-1.e^{-x}}{\sqrt{1-(e^{-x}{)}^2}}].dx$

$=ln|e^x+\sqrt{e^{2x}-1}|+si{n}^{-1}{e}^{-x}+c$