# 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 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:

Declare

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

Differential

*d**u*= (2*x*+ 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.