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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.


    Included Quiz Questions

    1. the technique only performs a part of the original integration
    2. it is the inverse of the product rule for differentiation
    3. the integrand is split into parts
    4. it is the inverse of the chain rule for differentiation
    5. it is partial fraction method
    1. ln(x)
    2. x
    3. xln(x)
    4. 1
    5. 1/x
    1. u= tan⁻¹(x) and v'=x
    2. u=x and v'=tan⁻¹(x)
    3. u=tan(x) and v'=x
    4. u=tan(x) and v'=x²
    5. u=x² and v'=tan⁻¹(x)
    1. (-x/6)cos(6x)+(1/36)sin(6x)+c
    2. xcos(6x)+(1/36)sin(6x)+c
    3. (x/6)cos(6x)+(1/36)sin(6x)+c
    4. 6xcos(6x)+(1/36)sin(6x)+c
    5. -6xcos(6x)+36sin(6x)+c

    Author of lecture Integration by Parts: Example 2

     Batool Akmal

    Batool Akmal


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