Video: EG19M1-DiffAndInt-Q12Bv3

∫ ln π‘₯ dπ‘₯.

01:36

Video Transcript

Find the integral of the natural log of π‘₯ dπ‘₯.

To solve this integral, we’re going to use integration by parts. The formula that we’ll use is the integral of 𝑒 times d𝑣 dπ‘₯ dπ‘₯ is equal to 𝑒 times 𝑣 minus the integral of 𝑣 d𝑒 dπ‘₯ dπ‘₯. So notice we only have the natural log of π‘₯. That can be 𝑒. And then we can actually take natural log of π‘₯ as natural log of π‘₯ times one. This way, one will represent d𝑣 dπ‘₯.

So we wanted to use a 𝑒 that we can take the derivative very easily and then take d𝑣 dπ‘₯ as something that we can take the entire derivative very easily. So taking the derivative of 𝑒 with respect to π‘₯, we get one over π‘₯, because the derivative of the natural log of π‘₯ is one over π‘₯. And then the antiderivative of one would be π‘₯. So 𝑣 would be equal to π‘₯.

So now let’s plug this into our formula. Notice we have 𝑒 times 𝑣, which would be the natural log of π‘₯ times π‘₯. However, written like this, you may be tempted to write natural log of π‘₯ squared, which is not true. So it’s safer to write it as π‘₯ times the natural log of π‘₯. And we have minus the integral of π‘₯ because that’s what 𝑣 is equal to. And then d𝑒 dπ‘₯ can be replaced with one over π‘₯ dπ‘₯.

With our integral, the π‘₯s cancel. And we have the integral of one. And the integral of one with respect to π‘₯ is π‘₯. Therefore, we have π‘₯ natural log of π‘₯ minus π‘₯. But don’t forget to add the plus 𝐢. Therefore, our final answer will be π‘₯ natural log of π‘₯ minus π‘₯ plus 𝐢.

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