Q66 / 150 · MHT CET 2021 (24 Sep Shift 2) · Mathematics
$\int \frac{d x}{e^x+e^{-x}+2}=$
- A$\frac{1}{\mathrm{e}^{2 \mathrm{x}}+1}+\mathrm{c}$
- Correct answerB$\frac{-1}{\mathrm{e}^{\mathrm{x}}+1}+\mathrm{c}$
- C$\frac{1}{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}$
- D$\frac{-1}{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}$
Solution
Let
$\begin{aligned}
I & =\int \frac{d x}{e^x+e^{-x}+2} \\
& =\int \frac{d x}{e^x+\frac{1}{e^x}+2}=\int \frac{e^x d x}{e^{2 x}+2 e^x+1}=\int \frac{e^x}{\left(e^x+1\right)^2} d x
\end{aligned}$
Put $\mathrm{e}^{\mathrm{x}}+1=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$I=\int \frac{d t}{t^2}=\int t^2 d t=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c=\frac{-1}{e^x+1}+c$
Hence, the correct answer is option: 'B'.