 Hello and welcome to the GVSU Calculus screencasts. In this edition, we will do an example of integration by parts. Recall that integration by parts undoes the product rule for differentiation, so the rule for integration by parts is the integral of Udv is Uv minus the integral of Vdu. So in this screencast, we're going to work through an example where we need to use the integration by parts formula several times to evaluate a single definite integral. So consider the problem of integrating x cubed plus x squared plus one e to the x dx. With U equal x cubed plus x squared plus one, we have du is 3x squared plus 2x dx. And then setting dv equal e to the x dx and integrating gives us v is e to the x. Applying the integration by parts formula gives us the integral of x cubed plus x squared plus one times e to the x dx is u times v. Now u remember is x cubed plus x squared plus one and v is e to the x minus the integral of v du. v again is e to the x and du is 3x squared plus 2x times dx. Now this new integral that we get, integral 3x squared plus 2x times e to the x dx, is less complicated than the one we started with, but it's still not one that's easy to determine. Now again the integrand is a product 3x squared plus 2x times e to the x dx. So let's try integration by parts again. And the same idea as whole. We're going to integrate by setting u equal to 3x squared plus 2x because differentiating the polynomial will make things easier and dv is e to the x dx. So du in this case is 6x plus 2 dx and v is again e to the x. So the integration by parts formula tells us the integral of x cubed plus x squared plus one times e to the x dx is what we found to be x cubed plus x squared plus one times e to the x minus this integral 3x squared plus 2x times e to the x dx. Now we apply the integration by parts formula to the integral of 3x squared plus 2x times e to the x dx and that gives us u times v and u is 3x squared plus 2x and v is e to the x minus the integral of v that's e to the x times du 6x plus 2 times dx. Now distributing the subtraction across the material in the parentheses shows that the integral of x cubed plus x squared plus one times e to the x dx is x cubed plus x squared plus one times e to the x minus 3x squared plus 2x times e to the x plus 6x plus integral of 6x plus 2 times e to the x dx. So to finish our integration we need to integrate 6x plus 2 times e to the x dx. Again it's the same problem we had a product so we might try integration by parts. Let's set u equal to the polynomial part 6x plus 2 and dv is e to the x dx. Then we have du is 6 dx and v is e to the x again. So we apply integration by parts one more time on this integral 6x plus 2 e to the x dx in blue and that gives us u times v, u is 6x plus 2 and v is e to the x minus the integral of v du, v is e to the x and du is 6 dx. So we end up with the integral of x cubed plus x squared plus one e to the x dx is reduced to x cubed plus x squared plus one e to the x minus 3x squared plus 2x times e to the x plus 6x plus 2 times e to the x minus the integral of 6 e to the x dx. To complete our integration process we just have to integrate 6 e to the x dx and we can do that one directly. We can factor out the 6 and the integral of e to the x is just e to the x. So we conclude that the integral of x cubed plus x squared plus one e to the x dx is x cubed plus x squared plus one e to the x minus 3x squared plus 2x times e to the x plus 6x plus 2 e to the x minus 6 e to the x plus c. And that concludes this example of integration by parts. We hope to see you again soon.