Q100 / 170 · Manipal (MET) 2018 · Mathematics

$\int \frac{2 d x}{\left(e^x+e^{-x}\right)^2}$ is equal to

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

Solution

$\text { Let } \begin{aligned} I & =\int \frac{2 d x}{\left(e^x+e^{-x}\right)^2}=\int \frac{2 d x}{e^{2 x}+e^{-2 x}+2} \\ & =\int \frac{2 e^{2 x} d x}{e^{4 x}+2 e^{2 x}+1} \end{aligned}$ Put $\quad e^{2 x}=t$ $\Rightarrow \quad 2 e^{2 x} d x=d t$ $\begin{aligned} \therefore \quad I & =\int \frac{d t}{t^2+2 t+1}=\int \frac{d t}{(t+1)^2}=-\frac{1}{t+1}+c \\ & =-\frac{1}{e^{2 x}+1}=-\frac{e^{-x}}{e^x+e^{-x}}+c\end{aligned}$

Hence, the correct answer is option: 'A'.

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