Question

$\int \frac{dx}{e^{x} + e^{-x} + 2 }$ is equal to

• A. $\frac{1}{e^{x} + 1 } + C$
• B. $\frac{-1}{e^{x} + 1 } + C$
• C. $\frac{1}{ 1+e^{-x} } + C$
• D. $\frac{1}{e^{-x} - 1 } + C$
• E. $\frac{1}{e^{x} - 1 } + C$

Answer 1

Answer: (B)

Let $I =\int \frac{d x}{e^{x}+e^{-x}+2} $
$=\int \frac{e^{x}}{e^{2 x}+2 e^{x}+1} d x$
Put $e^{x}=t$
$\Rightarrow e^{x} d x=d t$
$\therefore I=\int \frac{d t}{t^{2}+2 t+1}$
$=\int \frac{d t}{(t+1)^{2}}$
$=\frac{-1}{t+1}+C$
$=\frac{-1}{e^{x}+1}+C$