$\int \frac{d x}{e^x+e^{-x}+2}=$

$\int \frac{d x}{e^x+e^{-x}+2}=$
  1. $\frac{1}{\mathrm{e}^{2 \mathrm{x}}+1}+\mathrm{c}$
  2. $\frac{-1}{\mathrm{e}^{\mathrm{x}}+1}+\mathrm{c}$
  3. $\frac{1}{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}$
  4. $\frac{-1}{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}$

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 $\mathrm{e}^{\mathrm{x}}+1=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$ $I=\int \frac{d t}{t^2}=\int t^2 d t=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c=\frac{-1}{e^x+1}+c$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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