Substituting with Expressions of the Form f(x) Multiplied by h(g(x))
When g‘(x) = f(x), you can use the substitution u = g(x) to integrate expressions of the form f(x) multiplied by h(g(x)), provided that h is a function that you already know how to integrate.
Variable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration.
Here’s a hairylooking integral that actually responds well to substitution:
The key insight here is that the numerator of this fraction is the derivative of the inner function in the denominator. Watch how this plays out in this substitution:

Declare u equal to the inner function in the denominator and make the substitution:
Here’s the substitution:

Differential du = (2x + 1) dx:

The second part of the substitution now becomes clear:
Notice how this substitution hinges on the fact that the numerator is the derivative of the inner function in the denominator. (You may think that this is quite a coincidence, but coincidences like these happen all the time on exams!)

Integration is now quite straightforward:
You take an extra step to remove the fraction before you integrate:

Substitute back x^{2} + x – 5 for u:
Checking the answer by differentiating with the Chain Rule reveals how this problem was set up in the first place:
Here’s another example where you make a variable substitution:
Notice that the derivative of x^{4} – 1 is x^{3}, off by a constant factor. So here’s the declaration, followed by the differentiation:
Now you can just do both substitutions at once:
At this point, you can solve the integral simply.
Similarly, here’s another example:
At first glance, this integral looks just plain horrible. But on further inspection, notice that the derivative of cot x is –csc^{2} x, so this looks like another good candidate:
This results in the following substitution:
Again, this is another integral that you can solve.