[MUSIC] We could figure out our anti derivative for log x, by just guessing
and then checking our answer. But we can also get that anti derivative
by using integration by parts. So I want to know an anti derivative of ,
log x dx. And just to remind you, this is natural
log. At first, it looks like there's nothing
helpful here. What would my u and my dv be?
The trick is to sit u log x that's the part of the anagram I'm going to
differentiate as dv. To be dx and it gets quite surprising,
and that helps. Let's see, so if u is log x then du is 1
over x dx. And if dv is dx then an anti-derivative,
just be x. Now, what does parts tell us?
So by parts the integral of u dv, so log x dx, that's what I'm interested in.
Is u v so x log x minus the integral of vdu, which is x times du.
And I've got x log x minus x times 1 over x, that's just 1.
So I'm just anti-differentiating dx. Then I've got x log x, what's an
anti-derviative of 1, just x, and I'll add a constant.
So I'm claiming that the antiderivative of log x is x log x minus x plus some
constant. I think it's really surprising that, at
least in this case, dv equals dx is a great choice.
Well here's a similar integration problem that you can try the same kind of
technique on. For example, here's a challenge.
Can you find an anti derivative of log of x squared?