𝘶-substitution: logarithmic function

We are faced with a fairly daunting-looking indefinite integral of pi over x natural log of x dx. Now, what can we do to address this? Is u-substitution a possibility here? Well for u-substitution, we want to look for an expression and its derivative. Well, what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? du is going to be the derivative of the natural log of x with respect to x, which is just 1/x dx. This is an equivalent statement to saying that du dx is equal to 1/x. So do we see a 1/x dx anywhere in this original expression? Well, it's kind of hiding. It's not so obvious, but this x in the denominator is essentially a 1/x. And then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi. I should do that in a different color since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here. And we could have assumed that, from the get go, this original expression was only defined for positive values of x because you had to take the natural log here, and it wasn't an absolute value. So we can leave this as just a natural log of x, but this also works for the situations now because we're doing the absolute value of that where the natural log of x might have been a negative number. For example, if it was a natural log of 0.5 or, who knows, whatever it might be. But then we are all done. We have simplified what seemed like a kind of daunting expression.