Question Video: Integrating the Natural Logarithm Function by Using Integration by Parts | Nagwa Question Video: Integrating the Natural Logarithm Function by Using Integration by Parts | Nagwa

Question Video: Integrating the Natural Logarithm Function by Using Integration by Parts Mathematics

Integrate ∫In π‘₯ dπ‘₯ by parts using 𝑒 = In π‘₯ and d𝑣 = dπ‘₯.

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Video Transcript

Integrate the natural logarithm of π‘₯ with respect to π‘₯ by parts using 𝑒 is equal to the natural logarithm of π‘₯ and d𝑣 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 𝑒 times d𝑣 by dπ‘₯ with respect to π‘₯ is equal to 𝑒 times 𝑣 minus the indefinite integral of 𝑣 multiplied by d𝑒 by dπ‘₯ with respect to π‘₯.

Usually when we’re using integration by parts, we first need to determine which function we set as 𝑒 and which one we set as d𝑣 by dπ‘₯. However, we’re already told this information in the question. We’re told to set 𝑒 equal to natural logarithm of π‘₯. And we’re told in terms of differentials, d𝑣 is equal to dπ‘₯. This means that d𝑣 by dπ‘₯ is equal to one. We can then see if we set 𝑒 equal to the natural logarithm of π‘₯ and d𝑣 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, we’re going to need to find expressions for 𝑣 and d𝑒 by dπ‘₯. Let’s start by differentiating 𝑒 with respect to π‘₯. That’s the derivative of the natural logarithm with respect to π‘₯, which we know is the reciprocal function d𝑒 by dπ‘₯ is one over π‘₯. Since d𝑣 by dπ‘₯ is equal to one, the derivative of 𝑣 with respect to π‘₯ is equal to one. In other words, 𝑣 is an antiderivative of one. And we know the derivative of π‘₯ with respect to π‘₯ is one, so we’ll set 𝑣 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, we’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. It’s equal to π‘₯ plus the constant of integration 𝐢.

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 𝐢. 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.

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