Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form.