Question

$\int \frac{e^x}{4+e^{2x}}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)+C$
• B. ${\tan }^{-1}\left(\frac{e^{2x}}{2}\right)+C$
• C. $\frac{1}{2}{\tan }^{-1}\left(\frac{e^{2x}}{2}\right)+C$
• D. $\ln \frac{(e^x-2)}{(e^x+2)}+C$

Answer 1

The correct option is A $\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)+C$

we have,
$\int \frac{e^x}{4+e^{2x}}dx$
Let, $e^x=t\Rightarrow e^{x}dx=dt$
Integral becomes
$\int \frac{dt}{4+t^2}=\frac{1}{2}{\tan }^{-1}\left(\frac{t}{2}\right)+C=\frac{1}{2}{\tan }^{-1}\left(\frac{e^x}{2}\right)+C$