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Integration by Parts: Example 2

by Batool Akmal
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    00:01 Let’s have a look at our second example.

    00:04 We are now dealing with X to the power of 4 multiplied ln of 3X.

    00:10 Let’s write this out firstly, so I have X to the power of 4 and that is being multiplied by ln of 3X.

    00:19 Hopefully, I have had a little thought about substitution.

    00:22 This won’t really work with substitution because U would be 3X and you are not really going to make your function any easier when you use substitution.

    00:31 I am going to find my formula for integration by parts.

    00:34 So, uv' equals to uv minus the integral of vdu dx.

    00:41 Don’t tend to put the X’s there.

    00:44 And remember that this is the only part I’m doing once I’ve split it into uv'.

    00:49 Now, I’m going to show you a little mistake that we can make here before we start.

    00:54 Imagine now if I split this into U and V.

    00:57 So remember that this isn’t the correct way of doing it, I’m just going to show you something that you could do wrong and then how to correct ourselves.

    01:05 If I say U equals to x to the 4 just like we did previously because that how it’s written and V dashed equals to ln of 3x.

    01:14 This, I can differentiate, no problem, but do you know the integral of ln of 3x? You don’t.

    01:22 We know that ln differentiates to 1 over X, but we don’t know what it integrates to.

    01:27 We can divide the integral of this using integration by parts, but it’s a much longer expression.

    01:32 So in this case, I can stop and think, maybe this isn’t the right way for me, maybe integrating ln of x isn’t the right idea.

    01:40 I can differentiate ln of x.

    01:42 So in this case, I almost have no choice.

    01:44 I have to change my options.

    01:47 So I’m going to make U into ln of 3X and vx to the 4.

    01:52 And that’s only because I don’t have a choice.

    01:55 That’s the only way I can do this.

    01:57 So if I rewrite my V dashed here, so remember you are splitting it into U and V dashed.

    02:03 I have said now that U equals to ln of 3X and V dashed is X to the fourth.

    02:10 You may think that this is going to grow when we integrate this, the power of X to the fourth will become bigger, which it will.

    02:16 But once again, we don’t have an option here, so we are just going to go with this and hope that it becomes easier.

    02:22 Okay, remember what we do here.

    02:24 This, I’m going to differentiate and this, I’m going to integrate.

    02:27 In order to get from U to U dashed, I need to differentiate it.

    02:31 In order to go from V dashed to V, I’m going to integrate it, so I’m going backwards.

    02:36 U' here, ln of anything just goes to 1 over that thing.

    02:41 So if it’s ln of 3X, we get 1 over 3X, but don’t forget this is a function of a function.

    02:47 So you have a 3X on the inside, so you have to multiply it with the 3.

    02:52 If I tidy this up, I will get -- the 3’s can cancel out, so the answer to this is just 1 over X, nice and straightforward.

    03:00 I’m going to integrate my V dashed.

    03:03 So when I integrate this, you add 1 to the power and you divide by new power.

    03:08 Don’t worry about adding your plus C here.

    03:10 The C is just a constant, a number, that we can put right at the end of the integral.

    03:16 Putting it all into our formula, so I’m just going to use this part of the formula here.

    03:20 So I will say that UV.

    03:22 So here is my U and here is my V, so those 2 terms multiplied together firstly.

    03:28 X to the 5 over 5 multiplied by ln of 3x.

    03:32 I don’t have to do anything with this anymore, minus the integral.

    03:36 And now, my integral of vdu dx so it’s those 2 terms here.

    03:41 So I got V multiplied by dU by dX, so I have X to the 5 over 5 multiplied by 1 over x, dx.

    03:52 Okay.

    03:53 This perhaps to look at.

    03:55 At first instance isn’t good news because we still got 2 functions of X’s here that are multiplying together, but the nature of these functions of X’s makes it straightforward.

    04:05 Because if you think about it, it’s X to the 5 divided by X.

    04:09 You can cancel this X to get X to the 4.

    04:12 So, we can combine this into one function, which is what we want because those are the kind of functions that we can integrate.

    04:19 So if I rewrite this again to show you what we are left with, I have the integral of one of this X’s has canceled with one of the 5 at the top.

    04:29 So you have X to the 4 over 5 dx.

    04:32 And again, we have successfully simplified our integral.

    04:36 So rather than dealing with two functions of X here, X to the 4 times ln of 3x, we just have one function to solve to integrate.

    04:45 I can take this 5 out if this looks unpleasant.

    04:49 So, I have ln of 3X.

    04:52 Because the 5 is just a constant, I can write it like this and then I have x to the 4, dx.

    04:58 Now, I just have to integrate X to the 4, which we know is straightforward, add 1 to the power divided by new power.

    05:04 So, this stays x to the 5 over 5, ln of 3X.

    05:09 We don’t have to do anything there, minus 1 over 5, that stays.

    05:13 When you integrate X to the 4, that goes to X to the 5 over 5.

    05:17 So, add 1 to the power divided by new power.

    05:20 And lastly, we just simplify that and put a plus c at the end, x to the 5, ln of 3x minus this 5s at the bottom can multiply.

    05:29 So X to the 5 over 25 and then you can put a plus C at the end.

    05:34 If this was a definite integral and it had limits, all you do at this stage once you've integrated it is just put limits into this entire integral.

    05:42 So put X’s everywhere and then take away the lower limit.

    05:47 So now, it’s your turn to practice some of these questions using integration by parts.

    05:52 You might be tempted to use substitution on the last question, but just for the sake of practice, I would say do them all by integration by parts.

    06:00 Good luck and I will see you in the exercise lecture.


    About the Lecture

    The lecture Integration by Parts: Example 2 by Batool Akmal is from the course Advanced Integration.


    Author of lecture Integration by Parts: Example 2

     Batool Akmal

    Batool Akmal


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