### Video Transcript

Integrate the natural logarithm
of π₯ with respect to π₯ by parts using π’ is equal to the natural logarithm of
π₯ and dπ£ is equal to dπ₯.

In this question, we want to
evaluate an indefinite integral by using integration by parts. We can do this by first
recalling integration by parts tells us the indefinite integral of π’ times dπ£
by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral
of π£ multiplied by dπ’ by dπ₯ with respect to π₯.

Usually when weβre using
integration by parts, we first need to determine which function we set as π’ and
which one we set as dπ£ by dπ₯. However, weβre already told
this information in the question. Weβre told to set π’ equal to
natural logarithm of π₯. And weβre told in terms of
differentials, dπ£ is equal to dπ₯. This means that dπ£ by dπ₯ is
equal to one. We can then see if we set π’
equal to the natural logarithm of π₯ and dπ£ by dπ₯ equal to one, then in our
integration by parts formula we get the indefinite integral of the natural
logarithm of π₯ with respect to π₯.

So to apply this, weβre going
to need to find expressions for π£ and dπ’ by dπ₯. Letβs start by differentiating
π’ with respect to π₯. Thatβs the derivative of the
natural logarithm with respect to π₯, which we know is the reciprocal function
dπ’ by dπ₯ is one over π₯. Since dπ£ by dπ₯ is equal to
one, the derivative of π£ with respect to π₯ is equal to one. In other words, π£ is an
antiderivative of one. And we know the derivative of
π₯ with respect to π₯ is one, so weβll set π£ equal to π₯.

Now, we just substitute all of
these expressions into our integration by parts formula. We get the indefinite integral
of the natural logarithm of π₯ with respect to π₯ is equal to π₯ times the
natural logarithm of π₯ minus the indefinite integral of π₯ multiplied by one
over π₯ with respect to π₯. And we can simplify this; π₯
multiplied by one over π₯ is just equal to one. So, weβre just left with π₯
times the natural logarithm of π₯ minus the indefinite integral of one with
respect to π₯. And we can evaluate this
integral. Itβs equal to π₯ plus the
constant of integration πΆ.

Therefore, we were able to show
the indefinite integral of the natural logarithm of π₯ with respect to π₯ is
equal to π₯ times the natural logarithm of π₯ minus π₯ plus πΆ. And integration by parts was
really useful for helping us evaluate this integral because the derivative of
the natural logarithm of π₯ is a much simpler expression than its integral.