1 ( e x + e - x ) 2 ? d x =

Question

$\int \frac{1}{{(e^{x} + e^{- x})}^{2}} d x =$

Smart Achievers · Accepted Answer

(1)

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

Explanation:

$\int \frac{1}{{(e^{x} + e^{- x})}^{2}} d x = \int \frac{e^{2 x}}{{(e^{2 x} + 1)}^{2}} d x$

Put $e^{2 x} + 1 = t$ $\Rightarrow 2 e^{2 x} d x = d t,$

then it reduces to

$\frac{1}{2} \int \frac{1}{t^{2}} d t = - \frac{1}{2} \cdot \frac{1}{t} + C$

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

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