Heres what i got so far....

I got

= u^2(u^3 + 2)^1/2

= (u^5 + 2u^2)^1/2 (DID I DO IT RIGHT?)

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- December 11th 2012, 10:22 AMasilvester635Basic Integration question
Heres what i got so far....

I got

= u^2(u^3 + 2)^1/2

= (u^5 + 2u^2)^1/2 (DID I DO IT RIGHT?) - December 11th 2012, 10:30 AMebainesRe: Basic Integration question
- December 11th 2012, 10:34 AMasilvester635Re: Basic Integration question
i got this (u^2)(w^1/2) is that what you meant?

- December 11th 2012, 10:58 AMebainesRe: Basic Integration question
If , then , or . So

Can you take it from here? - December 11th 2012, 11:04 AMasilvester635Re: Basic Integration question
sorry im lost... this is not how i learned it....

- December 11th 2012, 11:46 AMskeeterRe: Basic Integration question
- December 11th 2012, 12:14 PMasilvester635Re: Basic Integration question
oh snappp yessss integration by substitution.... okay lets start again from scratch..... so i pick which one to differentiate which is

w = u^3 + 2

dw = 3u^2 + 0 = 3u^2du

1/3dw = u^2du

= √w (1/3du)

= 1/3(2/3u^3/2 + c)

= 2/9u^3/2 + c - December 11th 2012, 08:32 PMvoyanceRe: Basic Integration question
Well, I'll give it a shot for you:

So, we have integ((u^2)sqrt((u^3) + 2))du)

By substitution, you can make w = u^3 + 2 and if you get the derivative of that, you get dw = 3u^2 du

By substituting what you have back into the integral, you get integ(((u^2)sqrt(w))du)

Now, since you have a u^2 and your dw is 3u^2, you have multiply the inside of the integral by 3 and multiply the outside by 1/3.

(1/3)integ(((3u^2)sqrt(w)du)) -> (1/3)integ(sqrt(w)dw)

From here, you get the anti derivative -> (1/3)(2/3)(w^3/2) -> (2/9)(w^3/2) + c

c is a constant...

In other words, you got it right - December 12th 2012, 06:59 AMebainesRe: Basic Integration question
Well actually the OP didn't get it right because what he had for an answer was 2/9u^3/2 + c. In any event the next (and last )step is to replace w with u^3 +2, to yield . It's always a good idea to check the answer by taking the derivative to verify that it yields as expected.

- December 12th 2012, 07:15 AMasilvester635Re: Basic Integration question
i knew that i just didn't care to put it was pretty obvious that it was the last step.... but thank you