 So there's a couple of special cases of integration by parts that we'll take a look at. So let's try and find the integral of 3x squared e to minus x dx. So we'll want to split this integrand into a product of f of x and g prime of x. And so if I take f of x equals 3x squared, then g prime of x must be... Now this only works if I can find both the derivative of f of x and the anti-derivative of g prime of x. And we're able to find both of these, so we can proceed, and that gives us a new integral. But how do we evaluate this one? We'll take a cue from the tortoise and the hare. Sometimes we just have to keep working at it. We can do integration by parts a second time, where again we'll need to split up our integrand into a product of something we can differentiate and something we can integrate. And if we make these choices, we're able to get another integral, but this one we can evaluate directly, and then we know what this term is, so it could substitute in to get our final answer plus c. Another thing we might be able to do appears in problems like the integral of e to the x cosine x. So we need to split this into something we can integrate and something we can differentiate. So let's try these. Differentiating u, anti-differentiating dv, applying our integration by parts theorem, and we have another integral we have to do. So let's do integration by parts once again. This time we notice that if we let u equals sine x and dv equals e to the x, we actually just go back to our original problem. So let's use the choice u equals e to the x and dv equals sine x dx. Differentiating u, anti-differentiating dv, and applying integration by parts, so we can replace that in our integral, which leaves us with another integral we can evaluate by parts. And so we, wait a minute, this anti-derivative e to the x cosine x dx is exactly what we started with. So it looks like we've come in a big circle. Well, not quite. There is a way out of this circle because notice that this gives us an equation that we can use to solve for our integral. So if we take our equation and do a little bit of algebra, then solve for the integral, we find we do have an expression for it. And again, since this is an indefinite integral, we do want to include that constant of integration.