Question

integration (e^x+e^-x) ^2 . (e^x-e^-x) dx​

Answer 1

Answer:

∫dxex+e−x

Multiplying numerator and denominator by ex

∫exe2x+1dx

Put ex=t therefore dt=exdx

∫dt1+t2

=arctant+C

Putting in original value, final answer is

=arctan(ex)+C

Well, another way to do this is using some Complex analysis.

Using Euler’s Identity

eix=cosx+isinx

ex=cosxi+isinxi

ex=cos(ix)−isin(ix)

Similarly

e−x=cos(ix)+isin(ix)

Step-by-step explanation:

hope it's helpful