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We are faced with a fairly daunting looking

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indefinite integral of pi over x natural log of x dx.

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Now what could we do to address this?

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Is u-substitution a possibility here?

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Well, for u substitution, we want to look for an expression and its derivative.

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Well, what happens if we set u equal to the natural log of x?

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Now what would "d u" be equal to in that scenario?

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"du" is going to be the derivative of the natural log of x with respect to x

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which is just 1 over x d x.

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This is an equivalent statement to saying that d u d x is equal to one over x.

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So do we see the one over x anywhere in this original expression?

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Well, it's kind of hiding

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it's not so obvious, but this x in the denominator is essentially

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one over x and that's being multiplied by dx.

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Let me rewrite this original expression to make a little bit more sense.

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So the first thing I'm gonna do, is I'm gonna take

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the pi - I should do that in a different color

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since I've already used...

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let me take the pi and just take it out front.

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Imma just take the pi right in front of the integral.

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And so this is gonna become the integral of -

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and let me write the one over the natural log of x first.

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one over the natural log of x

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times one over x

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dx

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Now it becomes a little bit clearer.

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These are completely equivalent statements.

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But this makes it clear that yes,

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u substitution will work over here.

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We set our u equal to natural log of x.

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Then our d u is one over x d x.

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One over x d x.

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Our d u is one over x d x.

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Let's rewrite this integral.

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It's going to be equal to

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pi times the indefinite integral of

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one over u

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natural log of x is u

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we set that equal to natural log of x

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times du

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times du

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Now this becomes pretty straightforward.

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What is the antiderivative of all this business?

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And we've done very

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similar things like this multiple times already.

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This is going to be equal to

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pi

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times the natural log

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the natural log of the absolute value

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of u

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so that we can handle even negative values of u

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the natural log of the absolute value of u

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plus c

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ok, so we have a constant factor out here.

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plus c

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And we're almost done, we just have to un-substitute for the u.

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U is equal to natural log of x.

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So we end up with this kind of neat looking expression.

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The anti of this entire indefinite integral, we have simplified

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we have evaluated it

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and it is now equal to pi

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times the natural log

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of the absolute value of u

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but u is just the natural log of x

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the natural log of x

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and then we have this plus c

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right over here.

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And we could've assumed that from the get-go

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this original expression

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was only defined for positive values of x

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because you have to take the natural log here

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and there wasn't a absolute value.

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So we can leave this as just a natural log of x

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But this also works for the situations now

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cause we're taking the absolute value of that

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where the ln of x might have been a negative number.

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for example if it was a natural log of

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.5 or who knows whoever it might be.

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But we are all done.

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We have simplified what seemed

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like a kind of daunting expression.