# Help with a difficult integral please..

• Nov 9th 2009, 02:41 PM
ihatemath
Help 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

• Nov 9th 2009, 03:50 PM
Jameson
Quote:

Originally Posted by ihatemath
(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

Your substitution doesn't work because du is too messy. If you let u=x+3 then x=u-3 and du=dx. So this integral becomes $\frac{2u-7}{\sqrt{u}}du$. Break this up into two terms and use the power rule.
• Nov 9th 2009, 04:02 PM
ihatemath
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.
• Nov 9th 2009, 04:05 PM
ihatemath
After that, I get 4u^3/2-14u^1/2.. substitue it in.. it doesn't give me the correct anser.
• Nov 9th 2009, 04:07 PM
RhotelidlewildR
Quote:

Originally Posted by ihatemath
(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

Did you cancel your exponents after you took the integral of your substitution. Then all you have to do is backsub and evaluate for x
• Nov 9th 2009, 04:08 PM
RhotelidlewildR
Quote:

Originally Posted by ihatemath
After that, I get 4u^3/2-14u^1/2.. substitue it in.. it doesn't give me the correct anser.

((4u^(3/2)/(3)-14u^(1/2)
• Nov 9th 2009, 04:15 PM
ihatemath
sorry thats what i meant.. substitute u in.. it doesn't work. believe me. ive tried.
• Nov 9th 2009, 04:16 PM
Jameson
Quote:

Originally Posted by ihatemath
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.

$\int \frac{2u-7}{\sqrt{u}}du=\int \frac{2u}{\sqrt{u}}du-\int \frac{7}{\sqrt{u}}du$

$\int \frac{2u}{\sqrt{u}}du=2\sqrt{u}du=\frac{4u^{\frac{ 3}{2}}}{3}$

$\int -7u^{\frac{-1}{2}}du=-14\sqrt{u}$

Is this what you got?
• Nov 9th 2009, 04:21 PM
ihatemath
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)
• Nov 9th 2009, 04:22 PM
ihatemath
Oh, well looking at your post again.. yes that is what I got for the integrals once i separted them..
• Nov 9th 2009, 04:37 PM
Jameson
Quote:

Originally Posted by ihatemath
Oh, well looking at your post again.. yes that is what I got for the integrals once i separted them..

The final answer completely simplified is what you wrote earlier. I'm trying but can't get it into that form. The method I suggested so far should be right though.
• Nov 9th 2009, 04:44 PM
Jameson
$(2\sqrt{u})(\frac{2}{3}u-7)=(2\sqrt{u})(\frac{2u-21}{3})$