Question

$\int \frac{2dx}{{\left(e^x+e^{-x}\right)}^2}=$

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

Answer 1

The correct option is A

$-\frac{e^{-x}}{e^{-x}+e^x}+C$

Explanation for Correct answer:

Evaluating the given integral:

$$\begin{gathered} \int \frac{2}{{\left(e^x+e^{-x}\right)}^2}=\int \frac{2}{{\left(e^x+{\displaystyle \frac{1}{e^x}}\right)}^2}dx \\ =\int \frac{2}{{\displaystyle {\left(\frac{e^x{e}^x+1}{e^x}\right)}^2}}dx \\ =\int \frac{2e^x}{{\displaystyle {\left(e^{2x}+1\right)}^2}}dx \end{gathered}$$

Solving this by using the substitution method

Let $$\begin{gathered} t=e^{2x}+1 \\ \Rightarrow dt=2e^{2x}dx \end{gathered}$$

$$\begin{gathered} \int \frac{2}{{\left(e^x+e^{-x}\right)}^2}=\int \frac{dt}{t^2} \\ =-\frac{1}{t}+C \\ =-\frac{1}{e^{2x}+1}+C \\ =-\frac{e^{-x}}{e^{-x}+e^x}+C \end{gathered}$$

Hence, option (C) is the correct answer.