0
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Try

Integrate[ArcTan[E^x]/E^(2 x), x]//FunctionExpand //PowerExpand //Expand
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.