Main content

## Integrating with u-substitution

# 𝘶-substitution: logarithmic function

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.1 (EK)

## Video transcript

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.