(2x-1)/sqrt(x+3) dx

This is what I've tried

u=sqrt(x+3)

x=u^2-3

I then plugged x in for x and U in the appropriate place so I had

(2(u^2-3)-1)/U du

I have tried everything.. and nothing works.. please help. Thank You

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- November 9th 2009, 02:41 PMihatemathHelp with a difficult integral please..
(2x-1)/sqrt(x+3) dx

This is what I've tried

u=sqrt(x+3)

x=u^2-3

I then plugged x in for x and U in the appropriate place so I had

(2(u^2-3)-1)/U du

I have tried everything.. and nothing works.. please help. Thank You - November 9th 2009, 03:50 PMJameson
- November 9th 2009, 04:02 PMihatemath
I tried that, and it doesn't work. I do not get the correct answer.. believe me.. I've gone through half a notebook on this one problem.

- November 9th 2009, 04:05 PMihatemath
After that, I get 4u^3/2-14u^1/2.. substitue it in.. it doesn't give me the correct anser.

- November 9th 2009, 04:07 PMRhotelidlewildR
- November 9th 2009, 04:08 PMRhotelidlewildR
- November 9th 2009, 04:15 PMihatemath
sorry thats what i meant.. substitute u in.. it doesn't work. believe me. ive tried.

- November 9th 2009, 04:16 PMJameson
- November 9th 2009, 04:21 PMihatemath
No, I just substitued U in to the equation in my last post.. and its no where close to a resemblence of the answer. is what you have posted correct? Because thats not what I have for the answer. The solution I have is (2/3)(sqrt (x+3))(2x-15)

- November 9th 2009, 04:22 PMihatemath
Oh, well looking at your post again.. yes that is what I got for the integrals once i separted them..

- November 9th 2009, 04:37 PMJameson
- November 9th 2009, 04:44 PMJameson
Ok I got it finally. Combine the answers I had above and start with factoring.

Now plug back in for x.

:) - November 9th 2009, 05:14 PMihatemath
thank you!