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

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−12e2x+1+c

12e2x+1+c

−1e2x+1+c

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detailed solution

Correct option is A

I=∫1ex+e−x2dx=∫e2xe2x+12dx Put e2x+1=t⇒2e2xdx=dt⇒I=12∫1t2dt=−12⋅1t+c=−12e2x+1+c

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