 Hello, and welcome to another screencast about integration by parts. So now that you guys have gotten warmed up a little bit, I'm going to throw another multiple choice question at you. So here's your integration by parts formula. The indefinite integral we're going to look at now is the integral of x squared plus 1 times e to the x dx. And again, I put that word times in there, so I know we're definitely going to want to use integration by parts. Okay, so now my question to you is which one would be your best choice for you and DV? So pause the video, look over your four choices, maybe start doing some integration by parts and see if it works out nicely for you, and come back. All right, so welcome back. So hopefully after you look through your choices here, let's see which one is going to be the best, and I believe that is going to be choice D. Okay, why choice D? Well, let's go through and look at our other choices and see maybe why that's going to be a good one to eliminate. So in this function, we've got a polynomial, we've got this x squared plus 1, and then we've got the exponential function e to the x. Well, e to the x is easy to integrate and differentiate, so you know, that's fairly nice. But we want this polynomial to get smaller, not bigger. Okay, so definitely picking this polynomial for our u is going to make more sense. So looking at this one here, part a, or choice a, u being e to the x, there's nothing wrong with that, but dv being just x squared, well then, what about this plus 1? You know, you can't have any constants dangling in there, so that one doesn't make any sense, because it didn't take into account the whole function. B, so again, u is e to the x, that seems reasonable, but dv being x squared plus 1 dx. So if this is our choice for dv, that means we're going to be integrating that. So that means this polynomial is going to be getting bigger in degree, not smaller. That doesn't really make a whole lot of sense. Okay, choice c has kind of the same problems as choice a, so for this one I put u as our x squared, dv is e to the x. Well, dv being e to the x, dx is fine, you know, but again, u being just x squared, that doesn't make a lot of sense, because what do we do with that plus 1? Okay, so that's why choice d is definitely the best. Okay, so let's go ahead and go with choice d, so I'm going to set up my u over here as x squared plus 1, and my dv down here being e to the x, dx. And how you organize this is totally up to you, this is just kind of what I'm accustomed to doing, and then filling in the missing pieces that I need. Okay, so if my u is x squared plus 1, my du, so remember we're differentiating this direction, is 2x dx, and then if my dv is e to the x, dx, that means I want to integrate this way, but guess what? Integrating or differentiating e to the x doesn't matter, always going to give us e to the x. Okay, so you want to be happy if you see an integral with e to the x in it, because trust me, they're always going to be fairly nice to do. Okay, so let's put our pieces together then in our parts formula, so our original integral was x squared plus 1 times e to the x dx, so that's going to equal u, which is x squared plus 1 times v, which is e to the x, minus the integral of v, which is e to the x, du, which is 2x dx. Okay, so would you guys agree that we're starting off fairly good here? I think so, because our polynomial that we had in our original integral was a second degree, we still have a polynomial in this exponential function, but now it's a first degree, okay? So it's getting smaller, it's getting simpler, so that's pretty good, but we still have a product in here, e to the x times 2x dx, oh great, we're going to have to integration my parts again, okay, but that's all right, we need the practice. So I'm going to rewrite my first piece, and then just because I like to pretty things up a little bit, I'm going to pull this 2 out front of my integral, so I'm going to have minus 2, and then the integral of x e to the x dx, just so it kind of looks like the one I started with, not really. And then again, I'm going to pick my u's and my dv's the same way I did on the last one. So my dv is still going to be e to the x dx, my du, or my u is going to be x, so that makes my du dx, and my v again, e to the x, okay? So this is definitely a smart choice. If you have to do integration by parts multiple times, it's usually good to pick the same idea for your u and your dv. It may not be the exact same function, but as long as it's kind of the same general vicinity, that should be fine. Otherwise, if I would have picked my dv to be x, then I would have had to integrate that, and it would have gotten bigger, right? I don't want that, remember, you keep wanting to make things smaller. Okay, so putting these pieces together, and don't lose the other pieces that we started with. So x squared plus 1 e to the x minus 2. All right, I'm going to put in a big old parenthesis here. Okay, so now this integral x e to the x dx is what I've put my parts over here for, so uv, so that's going to be x times e to the x, and then minus the integral of v, which is e to the x, du, which is dx. Okay, so don't forget, let me highlight this piece in red. Don't forget about this negative 2 on the outside, though, so that's going to have to go to all of our pieces on here. Okay, so let me go ahead and rewrite this. No, I'm starting to run out of room, too. So we have x squared plus 1 e to the x. I'm going to go ahead and distribute that negative 2 through to both of these pieces. So I have negative 2x e to the x, and then plus the 2 times the integral of e to the x dx, and, again, this became a positive because I had a negative from the first integration by parts, and I had another negative from the second integration by parts, so that ended up giving me a positive. Okay, so now I have a function that's easy to integrate, one of our straightforward functions, so now I can write down my answer. So that's x squared plus 1 e to the x minus 2x e to the x, and then plus 2 e to the x, finally, plus c. So that is your final answer, and my challenge to you, oopsie, is to do the derivative of this function and convince yourself that you get back to where we started. Good luck.