$I=\int \frac{d x}{e^x+4 e^{-x}}=f(x)+c$ then f(x) is equal to

$\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)$

$I=\int \frac{d x}{e^x+4 e^{-x}}=f(x)+c$

$\Rightarrow I=\int \frac{e^x d x}{e^{2 x}+4}$

Let $e^{x}=t \Rightarrow e^{x} dx=dt$

$\Rightarrow I=\int \frac{d t}{t^2+4}=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+c=\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)+c$

Hence (2) is the correct answer.