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# C3 integration question Watch

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1. Integrate by parts: 2x(x-2)4 dx

I tried using the formula uv - integral (v - du/dx dx) but cant seem to do the second integral that I get?
2. (Original post by IDontKnowReally)
Integrate by parts: 2x(x-2)4 dx

I tried using the formula uv - integral (v - du/dx dx) but cant seem to do the second integral that I get?
It's

What have you done so far?
3. (Original post by NotNotBatman)
It's

What have you done so far?
Thats what I meant, my bad!

This is what I have:

(x-2)^4 * x^3 - integral (x^2 * 4(x-2)^3 dx)
4. (Original post by IDontKnowReally)
Thats what I meant, my bad!

This is what I have:

(x-2)^4 * x^3 - integral (x^2 * 4(x-2)^3 dx)
One of your terms are wrong. The bits in bold should be the same.
If you let then integrate to get

Then you can integrate by parts again a few times over.
5. (Original post by NotNotBatman)
Your terms are wrong. The bits in bold should be the same.
If you let then integrate to get
so v = x^2 ?

I still can't do the second integral though
6. (Original post by IDontKnowReally)
so v = x^2 ?

I still can't do the second integral though
Integrate by parts again. There will come a point when it's easier to expand than to do IBP.
7. (Original post by NotNotBatman)
Integrate by parts again.
I tried to integrate the second integral and got this:

1/3 * (x-2)3 * x3 - integral(x3 * (x-2)2 dx)

Now I cant integrate the one above
8. (Original post by IDontKnowReally)
I tried to integrate the second integral and got this:

1/3 * (x-2)3 * x3 - integral(x3 * (x-2)2 dx)

Now I cant integrate the one above
expand and integrate as normal. And you've factorised the 4 right?
9. (Original post by NotNotBatman)
expand and integrate as normal. And you've factorised the 4 right?
Oh I see, thank you.
how do you know when it is appropriate to expand and then integrate, and when to integrate by parts/substitution?
10. (Original post by IDontKnowReally)
Oh I see, thank you.
how do you know when it is appropriate to expand and then integrate, and when to integrate by parts/substitution?
When it's easier expand and integrate using add 1 to power and divide by new power rule. If the highest power is 2 of a composite function, then it'd probably be easier to expand, but if there were higher powers then use IBP. With a lot of practise, it becomes a bit more obvious.

I usually consider integration by recognition and then using a u - sub and then by parts.
11. (Original post by IDontKnowReally)
Oh I see, thank you.
how do you know when it is appropriate to expand and then integrate, and when to integrate by parts/substitution?
I think there are other methods for doing that question that are much simpler- if you do u and dv/dx the other way around in the first stage you only have to do integration by parts once, and no expanding brackets is needed.
12. (Original post by NotNotBatman)
When it's easier expand and integrate using add 1 to power and divide by new power rule. If the highest power is 2 of a composite function, then it'd probably be easier to expand, but if there were higher powers then use IBP. With a lot of practise, it becomes a bit more obvious.

I usually consider integration by recognition and then using a u - sub and then by parts.
Thats really helpful, thank you (it wont let me upvote you tho :/)
13. (Original post by IDontKnowReally)
Thats really helpful, thank you (it wont let me upvote you tho :/)
No worries as the poster above said, swapping the u and dv would make it easier, but it's good practice to try a number of methods, then you'll quickly notice the which method to use in the future.

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