. One way to do this would be to multiply the function out into
a fifteenth degree polynomial and use the basic rules, but this is not very
clever. Instead, substitute u = 2x + 1. Then, by taking derivatives, we
see that du = 2 dx or 
. So,
For example, suppose we wish to evaluate the integral of the function
. One way to do this would be to multiply the function out into
a fifteenth degree polynomial and use the basic rules, but this is not very
clever. Instead, substitute u = 2x + 1. Then, by taking derivatives, we
see that du = 2 dx or . So,
It commonly happens that du is not simply a constant times dx. In these
cases, we must be a little more careful. For example, let 
in the
following integral:
