# Question Video: Integrating the Natural Logarithm Function by Using Integration by Parts Mathematics • Higher Education

Integrate β«In π₯ dπ₯ by parts using π’ = In π₯ and dπ£ = dπ₯.

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### 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 weβre 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, weβre already told this information in the question. Weβre told to set π’ equal to natural logarithm of π₯. And weβre 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, weβre going to need to find expressions for π£ and dπ’ by dπ₯. Letβs start by differentiating π’ with respect to π₯. Thatβs 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 weβll 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, weβre 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. Itβs equal to π₯ plus the constant of integration πΆ.

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 πΆ. 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.