The following result of indefinite integral contains hypergeometric series, but the reference answer is $-\frac{1}{2}\left(e^{-2 x} \arctan e^{x}+e^{-x}+\arctan e^{x}\right)$.

Integrate[ArcTan[E^x]/E^(2 x), x, 
  GeneratedParameters -> C] // FullSimplify
D[Integrate[ArcTan[E^x]/E^(2 x), 
    x] - (-(1/2) (E^(-2 x) ArcTan[E^x] + E^-x + ArcTan[E^x])), 
  x] // FullSimplify

How can I further simplify the above indefinite integral result into the form of reference answer?


1 Answer 1



Integrate[ArcTan[E^x]/E^(2 x), x]//FunctionExpand //PowerExpand //Expand

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