Question

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

(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) $\frac{1}{e^x-e^{-x}}+C$

Answer 1

(a) $\frac{-e^{-x}}{e^x+e^{-x}}+C$

$$\begin{gathered} \mathrm{Let}I=\int \frac{2dx}{{\left(e^x+e^{-x}\right)}^2} \\ =\int \frac{2dx}{{\left(e^x+{\displaystyle \frac{1}{e^x}}\right)}^2} \\ =2\int \frac{e^{2x}dx}{{\left(e^{2x}+1\right)}^2} \\ \mathrm{Let}{e}^{2x}+1=t \\ \Rightarrow e^{2x}·2dx=dt \\ \Rightarrow e^{2x}·dx=\frac{dt}{2} \\ \therefore I=2\times \frac{1}{2}\int \frac{dt}{t^2} \\ =-\frac{1}{t}+C \\ =-\frac{1}{e^{2x}+1}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}t=e^{\mathit{2}x}+1\right) \end{gathered}$$

Dividing numerator and denominator by e^x
$$\begin{gathered} \Rightarrow I=\frac{-{\displaystyle \frac{1}{e^x}}}{e^x+{\displaystyle \frac{1}{e^x}}} \\ =\frac{-e^{-x}}{e^x+e^{-x}}+C \end{gathered}$$