Use the chain rule for derivatives (the derivative of f( g(x) ) = f'( g(x) ) * g'(x), or derivative of the outside times derivative of the inside). Integration by substitution is a little like the chain rule in reverse.
Identify f'( g(x) ) and g'(x) in the function you have to integrate. In other words, look at the function you need to integrate a little differently. For example, if your function is as follows:
∫ (10 x^4) / (20 + 2 x^5) dx
Notice that 10 x^4 is the derivative of 20 + 2x^5.
In this kind of relationship, create a new variable called u and set it equal to the part whose derivative appears elsewhere in the function. In this example, then, u = 20 + 2x^5.
Take the following function, for example:
∫ -2 e^-x dx
Say that u = -x, since although -2 is not the derivative of -x, it's close. Multiply the derivative of -x by 2.
Substitute u and u', which is the derivative of u, into your equation in place of the x-functions. In the example above, for instance, we would change our equation to look like this:
∫ u' / u dx
What happens if u' is off by some integer value? In that case, we still substitute in u and u', but we correct the equation by dividing or multiplying by that integer value as necessary. Let's look at the example above:
∫ -2 e^-x dx
We know that u = -x, so u' = -1. That means that -2 = u' * 2, so when we substitute we end up with an equation that looks like this:
∫ 2 u' (e^u) dx
Note that u' and dx together make du and rewrite the equation accordingly. In the two examples above, for instance, the result looks like this:
∫ 2 e^u du
∫ 1 / u du
Evaluate the integral using your basic rules of integration. If you don't remember all of them, look at the link under the Resources section below; it contains a table of common integrals. In the example problems, you would get the following results:
∫ 2 e^u du = 2 e^u
∫ 1 / u du = ln u
Substitute u's original value back into the equation in place of u. In the examples, you would end up with the following:
2 e^u = 2 e^-x
ln u = ln (20 + 2x^5)
Double-check your answer by taking the derivative to make sure it looks just like the original equation you had when you started the problem.