Integration by parts: ∫x²⋅𝑒ˣdx

Let's see if we can take the antiderivative of x squared times e to the x dx. Now, the key is to recognize when you can at least attempt to use integration by parts. And it might be a little bit obvious, because this video is about integration by parts. But the clue that integration by parts may be applicable is to say, look, I've got a function that's the product of two other functions-- in this case, x squared and e to the x. And integration by parts can be useful is if I can take the derivative of one of them and it becomes simpler. And if I take the antiderivative of the other one, it becomes no more complicated. So in this case, if I were to take the antiderivative of x squared, it does become simpler. It becomes 2x. And if I take the antiderivative of e to the x, it doesn't become any more complicated. So let's assign f of x to be equal to x squared. And we want that one to be the one where if I take the derivative it becomes simpler. Because I'm going to have to take the derivative of f of x right over here in the integration by parts formula. And let's assign g prime of x to be equal to e to the x. Because later, I'm going to have to take its antiderivative, and the antiderivative of e to the x is still just e to the x. So let me write this down. So we are saying that f of x-- I'll do it right over here-- f of x is equal to x squared, in which case f prime of x is going to be equal to 2x. And I'm not worrying about the constants right now. We'll just add a constant at the end to make sure that our antiderivative is in the most general form. And then g prime of x is equal to e to the x, which means its antiderivative, g of x, is still equal to e to the x. And now we're ready to apply the right-hand side right over here. So all this thing right over here is going to be equal to f of x, which is x squared-- and let me put it right underneath-- x squared times g of x, which is e to the x, minus. Let me do that in that yellow color. I want to make the colors match up-- minus the antiderivative of f prime of x. Well, f prime of x is 2x, times g of x. g of x is e to the x dx. So you might say, hey, Sal, we're left with another antiderivative, another indefinite integral right over here. How do we solve this one? And as you might guess, the key might be integration by parts again. And we're making progress. This right over here is a simpler expression than this. Notice we were able to reduce the degree of this x squared. It now is just a 2x. And what we can do to simplify this a little bit. Since 2 is just a scalar-- it's a constant, it's multiplying the function-- we can take that out of the integral sign. So let's take it this way. So let me rewrite it this way. We can only do that with a constant that's multiplying the function. So let me put the 2 right out here. And so now what we're concerned about is finding the integral-- let me write it right over here-- the integral of xe to the x dx. And this is another integration by parts problem. And so let's again apply the same principles of integration by parts. What, when I take its derivative, is going to get simpler? Well, x is going to get simpler when I take its derivative. So now, for the purposes of integration by parts, let's redefine f of x to be equal to just x. And then we can still have g prime of x equaling e to the x. And so in this case, let me write it all down. f of x is equal to x. f prime of x is equal to 1. g prime of x is equal to e to the x. g of x, just the antiderivative of this, is equal to e to the x. So let's apply integration by parts again. So this is going to be equal to f of x times g of x. Now f of x is x. g of x is e to the x, minus the antiderivative of f prime of x-- well, that's just 1-- times g of x-- e to the x. It's just 1 times e to the x dx. And remember, all I'm doing right now-- you might have lost track of things-- I'm just focused on this antiderivative. That antiderivative is that antiderivative there. If we can figure out what it is, we can then substitute back into our original expression. Now, you might appreciate integration by parts. What does this right over here simplify to? What is the antiderivative of 1 times e to the x dx? Or what is the antiderivative of 1 times e to the x? Well, it's just the antiderivative of e to the x, which is just e to the x. So this simplifies to x times e to the x minus the antiderivative of e to the x, which is just e to the x, so minus e to the x. And then we can take this and substitute it back. This is the antiderivative of this. So we can substitute it back up here to figure out the antiderivative of our original expression. So the antiderivative of our original expression-- we're getting really close-- is going to be equal to-- I'm going to use different colors so we can keep track of things. It's going to be equal to x squared times e to the x minus 2 times all of this business. So minus 2 times-- well, this antiderivative we just figured out is this. Minus 2 times xe to the x minus e to the x. And if we want, now is a good time to put our plus C. And of course, we can simplify this. This is equal to x squared. I like to keep the same colors. This is equal to x squared e to the x. You distribute the negative 2. You get minus 2xe to the x plus 2e to the x, and then finally, plus C. And we're done. We figured out the antiderivative of what looked like a kind of hairy-looking expression using integration by parts twice.