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Integration: Exercise 3

by Batool Akmal
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    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. And when we have those types of scenarios, we can bring it up and see what we can do with it. So my integral here is x over 5x plus 1 cube, dx. Integration by parts formula states this, uv' dx equals to uv minus the integral of vu' dx. So, we need every component. We need u, v, u' and v'. But before I do this, I’m just going to tidy this is up and make it more understandable for myself. So, I’m going to bring that up, 5x plus 1 to the minus 3, dx. And now, I can clearly see that it’s a product. So you have one function multiplying by the other function. If I say that u equals to x and v' equals to 5x plus 1 to the minus 3. This needs to differentiate. This needs to integrate. U' goes to 1 and V, now, be careful, we are integrating. When you integrate, add 1 to the power divided by new power divided by the differential. So we have 5x plus 1. 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. So, we get 5x plus 1 to the minus 2 over minus 10. Don’t take this down yet. 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. So put it all into...

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    The lecture Integration: Exercise 3 by Batool Akmal is from the course Advanced Integration.


    Author of lecture Integration: Exercise 3

     Batool Akmal

    Batool Akmal


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