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Math Help - Help with a difficult integral please..

  1. #1
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    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

    I have tried everything.. and nothing works.. please help. Thank You
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  2. #2
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    Quote Originally Posted by ihatemath View Post
    (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
    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.
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  3. #3
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    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.
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  4. #4
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    After that, I get 4u^3/2-14u^1/2.. substitue it in.. it doesn't give me the correct anser.
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  5. #5
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    Quote Originally Posted by ihatemath View Post
    (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
    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
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  6. #6
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    Quote Originally Posted by ihatemath View Post
    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)
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  7. #7
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    sorry thats what i meant.. substitute u in.. it doesn't work. believe me. ive tried.
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  8. #8
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    Quote Originally Posted by ihatemath View Post
    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?
    Last edited by Jameson; November 9th 2009 at 04:41 PM.
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  9. #9
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    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)
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  10. #10
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    Oh, well looking at your post again.. yes that is what I got for the integrals once i separted them..
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  11. #11
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    Quote Originally Posted by ihatemath View Post
    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.
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  12. #12
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    Ok I got it finally. Combine the answers I had above and start with factoring.

    (2\sqrt{u})(\frac{2}{3}u-7)=(2\sqrt{u})(\frac{2u-21}{3})

    Now plug back in for x.

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  13. #13
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    thank you!
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