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 hairy-looking 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:

1. Declare u equal to the inner function in the denominator and make the substitution:

Here’s the substitution:

2. Differential du = (2x + 1) dx:

3. 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!)

4. Integration is now quite straightforward:

You take an extra step to remove the fraction before you integrate:

5. Substitute back x2 + 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 x4 – 1 is x3, 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 –csc2 x, so this looks like another good candidate:

This results in the following substitution:

Again, this is another integral that you can solve.