Integration by Parts: Example 1

by Batool Akmal

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    00:01 And our last example which we are also going to do by integration by parts, but remember we are going to have to tweak it a little bit because you have a function as a denominator.

    00:09 And when we have those types of scenarios, we can bring it up and see what we can do with it.

    00:14 So my integral here is x over 5x plus 1 cube, dx.

    00:23 Integration by parts formula states this, uv' dx equals to uv minus the integral of vu' dx.

    00:40 So, we need every component.

    00:41 We need u, v, u' and v'.

    00:45 But before I do this, I’m just going to tidy this is up and make it more understandable for myself.

    00:49 So, I’m going to bring that up, 5x plus 1 to the minus 3, dx.

    00:54 And now, I can clearly see that it’s a product.

    00:56 So you have one function multiplying by the other function.

    01:00 If I say that u equals to x and v' equals to 5x plus 1 to the minus 3.

    01:08 This needs to differentiate.

    01:10 This needs to integrate.

    01:12 U' goes to 1 and V, now, be careful, we are integrating.

    01:17 When you integrate, add 1 to the power divided by new power divided by the differential.

    01:22 So we have 5x plus 1.

    01:24 Add 1, so, minus 3 plus 1 gives you minus 2, divided by new power and divided by the differential of the inside which is 5.

    01:34 So, we get 5x plus 1 to the minus 2 over minus 10.

    01:40 Don’t take this down yet.

    01:41 So, don’t take this power down as yet because we are still going to need to do some integration, some simplifying, so just leave it as it is and we will do that right at the end.

    01:51 So put it all into our formula, we have u times v, so we have that simplified version here.

    01:58 So I have X multiplied by 5x plus 1 to the minus 2 over minus 10, minus the integral of vdu/dx.

    02:08 So those two terms multiplied together.

    02:11 So that gives me 1 times 5x plus 1 to the minus 2 over minus 10.

    02:18 So in some sense, it’s good.

    02:19 We didn’t take this down because we are going to need it up as a numerator when we integrate.

    02:27 This, I can tidy up, this term here because we are not really doing anything more with this.

    02:32 So I can write this as minus X over 10.

    02:34 The 10 can move anywhere and I can also move that down, so I can write it as 5x plus 1 squared.

    02:41 I will tidy that up in the next step.

    02:43 We now need to integrate 5x plus 1 to the minus 2 dx, and the minus 10, I can take outside as 10.

    02:54 Because it was a minus, it makes this sign a plus, so it’s minus times a minus there, which will make it into a plus.

    03:01 So let’s integrate, this can stay as minus x over 10, 5X plus 1 to the power 2 plus 1 over 10.

    03:12 Integrate this now, remember what happens when you integrate, add 1 to the power divided by new power divided by the differential.

    03:20 So 5x plus 1 to the minus 1, so minus 2 plus 1 is minus 1 divided by minus 1 times by the differential of the inside which is 5x plus 1.

    03:32 I’ll put my plus C at the end once I have tidied this up a little bit.

    03:35 So I've got minus X over 10, 5x plus 1 squared and then this minus here will make this entire term minus.

    03:45 So imagine this minus is multiplying with this 10 which is then multiplying here.

    03:50 So this will make this entire term minus.

    03:54 The 10 and the 5 at the bottom can multiply to give you 1 over 50 and then I have 5x plus 1 to the minus 1 which I can bring down as 5x plus 1 to the positive 1 and I will put plus C at the end.

    04:09 So we have integrated two functions that are dividing each other using integration by parts just by bringing it up.

    04:17 A few last tips, just remember that try or think about substitution before.

    04:23 If you think substitution will make it a little bit more complicated, then move to integration by parts.

    04:29 Always look at your u' and v', so your first choices here are fairly important.

    04:35 It is common to get them wrong sometimes, but the only thing that you need to be careful about is when you have ln's.

    04:42 Because an ln function should never be your v' function, it should always be u, because we don’t know the integral of ln of x.

    04:50 You can do it.

    04:51 You can do it using integration by parts, but it gives a much longer and complicated result.

    04:57 So only when you have a function and ln, in that case make sure that ln becomes your u and everything else becomes your v'.

    About the Lecture

    The lecture Integration by Parts: Example 1 by Batool Akmal is from the course Advanced Integration.

    Included Quiz Questions

    1. ∫ f(x) g(x) dx
    2. ∫ f(x) + g(x) dx
    3. ∫ f(x) - g(x) dx
    4. ∫ f(g(x)) dx
    5. ∫ f(x) / g(x) dx
    1. ∫ uv' dx = uv - ∫ vu' dx
    2. ∫ uv' dx = uv + ∫ vu' dx
    3. ∫ uv' dx = uv - ∫ vu dx
    4. ∫ uv' dx = uv + ∫ vu dx
    5. ∫ uv' dx = uv - ∫ v'u dx
    1. u = x³ and v' = sin(x)
    2. u = x³ sin(x) and v' = x
    3. u = x³ and v' = 1
    4. u = x³ and v' = cos(x)
    5. u = x² and v' = x sin(x)
    1. { [2x - 1] [e^(2x) / 4] } + c
    2. { [2x - 1] [e^(x) / 4] } + c
    3. { [2x - 1] [e^(x) / 2] } + c
    4. { [2x - 1] [e^(2x) / 2] } + c
    5. { [x - 1] [e^(2x) / 2] } + c

    Author of lecture Integration by Parts: Example 1

     Batool Akmal

    Batool Akmal

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