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# C3 Integration Question watch

1. Okay, the question is integrate (2x + 3)^1/2 dx.

why?

I understand why the power is raised to 3/2 as you raise the power by 1 with integration, but where has the 1/3 come from? Can anybody tell me? Cheers!
2. u=whatever

Differentiate this. Put in terms of du. Boom. You can do the rest.

3. The 3/2 comes from the power and the 2 from the derivative of what's inside the bracket. See it?
4. You can use the inverse chain rule.
When you differentiate: , where u is a function of x, goes to

So when you integrate: goes to , where u' is the derivative of u.
5. I still don't completely understand, but I see where you're all going, I'll just need to do some more work on the chain rule and integration in general. Cheers
6. (Original post by Cowboy Dan)
I still don't completely understand, but I see where you're all going, I'll just need to do some more work on the chain rule and integration in general. Cheers
What? I could have sworn it was much easier than that.

Isn't it:

Power goes up by one, divide by new power times differential of the bracket. So,

EDIT:
Ah, yes. I am right just you've written it dodgy. You've put times 1/3 rather than putting the whole equation over 3.
7. (ax+b)^n = 1/a(n+1) (ax+b) ^ n+1

Can't be bothered with latex but that's the general rule.

8. (Original post by *MJ*)
(ax+b)^n = 1/a(n+1) (ax+b) ^ n+1

Can't be bothered with latex but that's the general rule.

No it isn't. It's the differential, not just the coefficient of X.
9. The rule for integrating brackets like that is
if (ax+b)^n
then the integral is 1/a x 1/(n+1) x (ax+b)^n+1
so you get 1/2 x 1/(3/2) x (2x+3)^3/2
and the 1/2 x 1/(3/2) works out to 1/3
hope that helps I'm majorly revising C3 atm too..
10. (Original post by RollerBall)
No it isn't. It's the differential, not just the coefficient of X.
I know, just answering OP's Q.

11. (Original post by *MJ*)
I know, just answering OP's Q.

Okay, was just making sure. I think it's better to correct somebody and risk them thinking you've got a huge ego than let somebody miss some marks in an exam.

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Updated: January 19, 2010
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