Question

integration of 1/y+1

Answer 1

 solution

∫1/(y+1)dy

by letting u=y+1    ⇒du/dy=1   ⇒du=dy

⇒= (∫1/u)du

⇒= logu + c

but u=y+1

⇒∫1/(y+1)dy     =log(y+1) + c