Evaluate the following integrals:
First of all, take ex common from the denominator, so we get
⇒
Assume e - x + 1 =t
d(e - x + 1) = dt
⇒ - e - xdx= dt
⇒ e - xdx= - dt
Substituting t and dt we get
But t =(e - x + 1)
⇒ .