### Video Transcript

Determine the integral from one to
π of the natural logarithm of π₯ divided by π₯ with respect to π₯.

Weβre given a definite integral
which weβre asked to evaluate. We can see our integrand is the
natural logarithm of π₯ over π₯. And this is not easy to
integrate. We donβt know an antiderivative for
this expression. And this is the quotient of two
functions. And we donβt know how to integrate
this directly. So, weβre going to need to find
some other way of integrating this expression.

We might be tempted to use
integration by parts. However, in this case, weβll see
that integration by substitution is easier. And the reason integration by
substitution will be easier is to take a look at our numerator, the natural
logarithm of π₯. If we were to try integrating this
by using the substitution π’ is equal to the natural logarithm of π₯, then by
differentiating both sides of this equation with respect to π₯, we would see that
dπ’ by dπ₯ is one over π₯.

And this is a useful result. Since one over π₯ appears in our
integrand, it will help simplify our integral. So, weβre ready to start using our
substitution. First, although we know dπ’ by dπ₯
is not a fraction, when weβre using integration by substitution, it can help to
think of it a little bit like a fraction. This gives us the equivalent
statement in terms of differentials dπ’ is equal to one over π₯ dπ₯.

Next, remember, weβre evaluating a
definite integral by using integration by substitution. So, we need to rewrite our limits
of integration. We do this by substituting our
limits of integration into our expression for π’. To find our new upper limit of
integration, we substitute π₯ is equal to π into this expression. This gives us π’ is equal to the
natural logarithm of π, which we know is equal to one. And we do the same to find our new
lower limit of integration. We substitute in π₯ is equal to
one, giving us π’ is equal to the natural logarithm of one, which we know is equal
to zero.

So, weβre now ready to rewrite our
integral by using our substitution. First, we showed the new limits of
integration would be lower limit zero, upper limit one. Next, weβll replace the natural
logarithm of π₯ with π’. And then finally, we know that one
over π₯ dπ₯ is equal to dπ’. So, weβve rewritten our integral as
the integral from zero to one of π’ with respect to π’.

And we can just evaluate this
integral by using the power rule for integration. The integral of π’ with respect to
π’ is π’ squared over two. Then, all we do is evaluate this at
the limits of integration, giving us one squared over two minus zero squared over
two, which we can calculate is just equal to one-half. Therefore, by using the
substitution π’ is equal to the natural logarithm of π₯, we were able to show the
integral from one to π of the natural logarithm of π₯ over π₯ with respect to π₯ is
equal to one-half.