 Hello and welcome to the screencast on integration by substitution or U substitution. Today we're going to be integrating this function, natural log of x over x. Now this one does not look like a good candidate for integration by substitution right off the bat. But there are a few times when substitutions aren't obvious, but they turn out to work really well. And it's always worth trying to find a substitution if you think it's even slightly possible. So this function does not look like it's composite and being composite is what the chain rule and so integration by substitution are all about. But there's one thing that I notice about this and that is I remember from calc one that the derivative with respect to x of the natural log of x is one over x. And the reason that's good is I remember for integration by substitution I'm looking for a function multiplied by its derivative. That's what I need to find in the integrand. And it just turns out that if I rewrite this integral by splitting that fraction apart I find a function natural log of x and its derivative one over x and that's exactly what I'm looking for. So taking a little bit of time to rewrite things using some algebra is well worthwhile. So now that I've identified this this tells me what my u and my du need to be. My u is the original function natural log of x and my du is the derivative of that function one over x with respect to x. And now that I have that I am already to substitute. In fact it's a very easy substitution natural log is u and the one over x dx is du and this is an easy integral to do. It's just a polynomial and this is one of the big things that integration by substitution can do. It can turn very complex difficult to deal with integrals into polynomials sometime. So I know how to integrate this. It's one half u squared plus an arbitrary constant. And since I started with x's I'm going to end with x's. So I'm going to replace the u with natural log of x and to write that square I'm going to be very careful. And there we go. That's the antiderivative. You can always check this by taking the derivative of the result. And if you notice to do this derivative you would have to use the chain rule. But when you did the chain rule this two would come down front and cancel out the one half and then disappear. And the resulting function wouldn't look like a composite function anymore. That's why what we had when we started didn't look very composite. So the lesson to take from this is it's worth doing a little bit of algebra and remember that you're looking for a function and its derivative multiplied together. If you can find that then you can use integration by substitution.