 Even though we'll spend quite a bit of time on various methods of integration, there's really only two integration techniques, U-substitution and integration by parts. Everything else is an algebraic or trigonometric identity or some application of U-substitutions or integration by parts. Integration by parts comes from the product rule. If from the product rule we know that the derivative of a product is going to be, and so the fundamental theorem of calculus says that if I integrate this mess, I get the original product, plus the constant of integration, which I won't worry about here for reasons that'll become apparent in a second. Now the additivity of the integral allows us to split this up, and we can solve for the first integral where the constant of integration that we should have included back here is going to be included when we find the anti-derivative of this. And that gives us a useful relationship, let f of x and g of x be functions, then the anti-derivative of f of x g prime of x is f of x g of x minus g of x f prime of x. In effect, what we're doing is we're switching the function that is differentiated. Alternatively, we often express the integration by parts rule as the integral of u dv is uv minus the integral of v du. This theorem suggests what's called integration by parts, and we'll proceed as follows. Let's say we want to find the integral of a product of functions. We'll choose one of the factors as our f of x, and let the other factor be our g prime of x. If it's even possible, or at least if it makes for a simpler integral, find the integral of g of x f prime of x. If it's not possible to do that, make a different selection. If it is possible, but it makes the integral uglier, use a different selection. And importantly, don't worry about the constant of anti-differentiation until the very end of the problem. That way we don't have too many constants cluttering up our work. So let's take a look at the integral of x sin of x. So if we want to use integration by parts, we have to view this as a product of two things, one of which will be f of x, and the other of which will be g prime of x. So we'll try it out. How about f of x equals x, and g prime of x equals sin of x? Now part of what will make this possible is we'll be able to find both f prime of x and g of x. Well, f prime of x is easy, because we know how to differentiate. g of x is going to be a little bit more complicated because that requires us to find an anti-derivative, and we might not be able to do that. But let's try anyway. And in this case, we're in luck. That's minus cosine of x, where again we're not going to worry about the constant of integration until the very end of the problem. So if we let f of x equals x, and g prime of x equals sin of x, then the integration by parts theorem tells us that our integral of x sin of x is going to be f of x times g of x minus the integral of g of x f prime of x. And we have an unresolved anti-derivative, and don't forget that constant of integration. What about the integral of x log of x? Well this time we'll let u equals x, and dv equals log x dx. So in this notation, u is the function that we'll be differentiating, and dv is the function that we'll be integrating. So we know that du equals dx, and v is a function whose derivative is log x, which we don't know. So let's try a different choice. So this time let's try u equals log x, and so dv has to be whatever's left over x dx. We need the derivative, and we need the anti-derivative of x dx. Since we can find both of them, it's possible to proceed. So the integral of u dv is going to be uv minus the integral of v du. We'll do a little bit of algebraic cleanup. We have an unresolved integral, so let's take care of that, and we get our final answer, plus c.