the integral of ∫ e^r/1+e^r?

 ∫  e^r  / [ 1 + e^r]   dr  =   Ln [ 1 + e^r ] + C

The key  here is to recognize  that  the derivative of 1 + e^r   =   e^r

So  remember that  if  u = 1 + e^r  then

 ∫   u '  /  u     du   =   Ln u + C