. If we try the
substitution 
, we are left with du = (2x + 6) dx. That
just doesn't help us any. However, if we complete the square on the denominator, we find that
Suppose we wanted to evaluate . If we try the
substitution , we are left with du = (2x + 6) dx. That
just doesn't help us any. However, if we complete the square on the denominator, we find that
The integral is now in a form compatible with trig substitution by
letting 
so that the integral is
