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

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

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

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- {\frac{1}{(e^{x} + 1}}^{2} + C

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- \frac{1}{(e^{x} + e^{- x})^{2}} + C

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Solution

The correct option is A $- \frac{e^{- x}}{(e^{x} + e^{- x})} + C$
Let $I = \int \frac{2 d x}{(e^{x} + e^{- x})^{2}} = \int \frac{2 e^{x}}{e^{x} (e^{x} + e^{- x})^{2}} d x$
Substitute $e^{x} = t \Rightarrow e^{x} d x = d t$
$∴ I = \int \frac{2 t}{(t^{2} + 1)^{2}} d t = - \frac{1}{1 + t^{2}} + C = - \frac{e^{- x}}{e^{x} + e^{- x}} + C$

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