Question

Solve:

$\int \frac{\sqrt{e^x}}{\sqrt{e^{-x}-e^x}}dx=?$

• A. ${\sin }^{-1}{e}^x+c$
• B. $2{\sin }^{-1}{e}^x+c$
• C. $-{\sin }^{-1}{e}^x+c$
• D. $-2{\sin }^{-1}{e}^x+c$

Answer 1

The correct option is A ${\sin }^{-1}{e}^x+c$

Let $I=\int \frac{\sqrt{e^x}dx}{\sqrt{e^{-x}-e^x}}=\int \frac{\sqrt{e^{2x}}dx}{\sqrt{1-e^{2x}}}=\int \frac{e^{x}dx}{\sqrt{1-e^{2x}}}$

Let $e^x=z\Rightarrow e^{x}dx=dz$

So, $I=\int \frac{dz}{\sqrt{1-z^2}}={\sin }^{-1}(z)+C={\sin }^{-1}{e}^x+C$