Imisstime

2.{ [(e^x)-(e^-x)]/[(e^x)+(e^-x)]} x (e^x)
give (e^2x-1)/(e^2x+1)
Since (e^2x+1) is 份母

Try d/dx {ln(e^2x+1)}
and u will get  (2e^2x)/(e^2x+1)

Different between 2e^2x/(e^2x+1) and [(e^x)-(e^-x)]/[(e^x)+(e^-x)]
is given as  [(e^x)-(e^-x)]/[(e^x)+(e^-x)] - {2e^2x/(e^2x+1)}
which is -(e^2x+1) / (e^2x+1) = -1
Therefore∫ [(e^x)-(e^-x)]/[(e^x)+(e^-x)] dx =  ln(e^2x+1) +∫ -1 dx

Imisstime

This is the way we are taught by the lesson...But they let u = something , and than du/dx
and work out the different between the du/dx and the question