First make a substitution and then use integration by parts to evaluate the integral. ∫x ln(1+x)

step 1 Hello and welcome. We are looking at chapter 5, section 5, problem 25. So we're given an integra

First make a substitution and then use integration by parts...

Key Concepts

Substitution Method

The substitution method involves changing the variable of integration to simplify the integrand. By selecting an appropriate substitution, a more complicated expression can be transformed into a simpler one, making the integral easier to evaluate. This technique is especially useful when the integrand contains a composite function or when one function’s derivative is present elsewhere in the integrand.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is used to integrate the product of two functions by transferring the derivative from one function to the other. This method is particularly effective when one of the functions simplifies upon differentiation and the other is easily integrable after the substitution.

Transcript

00:01 Hello and welcome.

00:02 We are looking at chapter 5, section 5, problem 25.

00:10 So we're given an integral, and we're given the instructions that we want to make a substitution and then use integration by parts.

00:23 So the substitution, i'm pretty sure we want to make, and i'll do a t substitution here.

00:30 We want to get rid of this 1 plus x.

00:31 That's the most logical one i can think of.

00:34 So if we do t equals 1 plus x, then d t is going to be dx, which is pretty convenient.

00:46 And we don't have to fool around too much with that anymore.

00:50 However, we do notice that x is going to be t minus 1.

00:59 If we're solving for x, so we're going to need to do that for this term here.

01:02 So if we make that substitution, we have the integral of t minus 1 times natural log of t d t.

01:22 So i kind of want to distribute that natural log of t in and split this into two integrals.

01:31 So we have integral of t, natural log of t, d t, and then minus the integral of natural log of t d t so we can use we can actually use integration by parts on both of these you could also look back at some previous problems where you had to take the integral of natural log of t and use that to help you get a general form however i'm going to show you integration by parts for both of these so we want you to equal t do we though i don't think we do because then we'd have to take the anti -derivative of natural log.

02:33 So let's change gears and make u, make you natural log of t, and then dv into t.

02:52 So then d -u would be 1 over t, d -t, and then v would be, oh, don't forget my d -t here, v would be t squared over two.

03:09 All right.

03:10 So let's tackle this first one, and we'll handle that second one in a bit.

03:19 So that would equal, just the first one here, natural log of t times t squared, that's all over two.

03:40 So that's the uv.

03:42 We want to minus it, subtract, i should say, the integral of v, so that would be t squared over two times 1 over t t t.

03:58 So working with that, put the t squared in front here.

04:06 That's all over two.

04:09 And then we subtract, this will be the, this ends up being one half the integral of t.

04:19 Once we simplify it.

04:30 So then this is minus one half times t squared over two.

04:37 And so we'll add the plus c, but we're going to have to add one at the end of the whole thing anyway.

04:44 So let's circle this as a partial answer, and we'll come back to it.

04:50 Let's take care of the second part here.

05:03 So for this one, we want u to be natural log of t, and we want dv to just be d t.

05:12 So in this case, du equals 1 over t dt, and then v is going to equal t...

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