∫(dx)/(eˆx + eˆ(-x))=∫(eˆx)/(1 + eˆ2x) t=eˆx x=ln (t) (dt)/(dx)=t => dx=1/t dt
∫t/(tˆ2 + 1) * 1/t dt => ∫1/(tˆ2 + 1) dt
Yeah! I solved it. You're right Dick, it isn't wrong at all. I just misthough there.
The next phase is using a formula ∫1/(xˆ2 + 1) dx = arctan x + C
So the...