Sometimes you need to integrate the product of a function (x) and a composition of functions (for example, the function 3x2 + 7 nested inside a square root function). If you were differentiating, you could use a combination of the Product Rule and the Chain Rule, but these options aren’t available for integration.



here’s how you integrate, step by step, using variable substitution:

  1. Declare a variable u as follows and substitute it into the integral:

    Here, you assign a value to u: let u = 3x2 + 7. Now substitute u into the integral:


    Make one more small rearrangement to place all the remaining x terms together:


    This rearrangement makes clear that you still have to find a substitution for x dx.

  2. Now differentiate the function u = 3x2 + 7:

    This gives you the differential,

  3. Substitute du/6 for x dx:


    You can move the fraction 1/6 outside the integral:

  4. Now you have an integral that you know how to evaluate.

    This example puts the square root in exponential form, to make sure that you see how to do this:

  5. To finish up, substitute 3x2 + 7 for u:


You can now check your integration by differentiating the result:


As if by magic, the derivative brings you back to the function you started with.