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Thread: root integral question..

  1. #1
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    root integral question..

    $\displaystyle
    \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=
    \int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}
    $
    i got read of one root but instead i got another one
    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    $\displaystyle
    \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=
    \int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}
    $
    i got read of one root but instead i got another one
    ??
    If you let $\displaystyle x = u^6$ then

    $\displaystyle \int \frac{6\,du}{u(u+1)(2u^2-u+1)}$
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  3. #3
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    when i substituted i didnt get your expression
    i got
    $\displaystyle
    \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)}
    $
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    when i substituted i didnt get your expression
    i got
    $\displaystyle
    \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)}
    $
    I'm sure you see the $\displaystyle u^5$ cancels and that if $\displaystyle f(u) = 2u^3+u^2+ 1$ then $\displaystyle f(-1) = 0 $ so $\displaystyle u+1$ is a factor and

    $\displaystyle 2u^3+u^2+ 1 = (u+1)(2u^2-u+1)$
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  5. #5
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    i cant solve this last part:
    $\displaystyle
    \frac{1}{3}\int\frac{-27u+2}{2u^2-u+1}
    $

    even when i split it:
    and take t=2u^2 -u +1

    ??
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    i cant solve this last part:
    $\displaystyle
    \frac{1}{3}\int\frac{-27u+2}{2u^2-u+1}
    $

    even when i split it:
    and take t=2u^2 -u +1

    ??
    You should get $\displaystyle
    \int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}
    $

    You have to split into 2 parts

    $\displaystyle \int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1} = -\frac{3}{8}\int\frac{4u-1}{2u^2-u+1} - \frac{1}{8}\int\frac{1}{2u^2-u+1}$

    The first part is integrable in log
    The second part is integrable in Arctan after setting t=u-1/4 then $\displaystyle v = \frac{4t}{\sqrt{7}}$
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  7. #7
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    i cant get your last part
    how you got this coefficients??

    2A+2B+c=0
    A-B+c+d=0
    B=-D

    A=6

    i got these equation
    and i cant get your coefficients
    ??
    Last edited by mr fantastic; Feb 20th 2009 at 04:19 AM. Reason: Merged posts
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  8. #8
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    $\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{1}{u} - \frac{\frac{1}{4}}{u+1} + \frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}$

    I forgot the "6" as numerator ... sorry
    Multiply everything above by 6
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  9. #9
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    its still does give the last part as i got it

    where did my coefficientcient equations differ yours??
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  10. #10
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    did i get the coefficient equations ..

    $\displaystyle
    \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\
    $
    $\displaystyle
    A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\
    $
    $\displaystyle
    u^3:\\
    $
    $\displaystyle
    2A-2B+C=0\\
    $
    $\displaystyle
    u^2:\\
    $
    $\displaystyle
    A+B+C+D=0\\
    u:\\
    $
    $\displaystyle
    -B+D=0\\
    $
    1:
    $\displaystyle
    A=1
    $
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  11. #11
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    Quote Originally Posted by transgalactic View Post
    $\displaystyle
    \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\
    $
    $\displaystyle
    A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\
    $
    $\displaystyle
    u^3:\\
    $
    $\displaystyle
    2A-2B+C=0\\
    $
    $\displaystyle
    u^2:\\
    $
    $\displaystyle
    A+B+C+D=0\\
    u:\\
    $
    $\displaystyle
    -B+D=0\\
    $
    1:
    $\displaystyle
    A=1
    $
    The denominator factorises to $\displaystyle u(1+2u^3+u^2) = u(u+1)(2u^2 - u +1)$.

    Thus, The fraction becomes $\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} + \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$.

    You have a sign error. $\displaystyle B$ is $\displaystyle +$ve, and not $\displaystyle -$ve. Your formula produced are correct, just a sign error has made them slightly differ.

    You can continue to use gaussian elimination after you've made the correction.
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  12. #12
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    Your equations are correct.

    There is a much easier way :

    $\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{1}{u(u+1)(2u^2-u+1)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$

    Multiply both sides by u

    $\displaystyle \frac{1}{(u+1)(2u^2-u+1)} = A - \frac{Bu}{u+1} + \frac{u(Cu+D)}{2u^2-u+1}$

    And set u to 0
    $\displaystyle 1 = A$

    Multiply both sides by u+1
    $\displaystyle \frac{1}{u(2u^2-u+1)} = \frac{A(u+1)}{u} - B + \frac{(u+1)(Cu+D)}{2u^2-u+1}$

    And set u to -1
    $\displaystyle -\frac{1}{4} = - B$
    Last edited by mr fantastic; Feb 21st 2009 at 04:20 AM. Reason: Required editing because it's been merged from another thread
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  13. #13
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    i tried again

    here is my full way till i got the coefficients
    where did i go wrong
    ??
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  14. #14
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    C=-9 is OK
    then -3+2D=0 gives D=3/2 and not 2/3
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  15. #15
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    $\displaystyle
    \int \frac{-9u+\frac{3}{2}}{2u^2-u+1}
    $

    but then the derivative of the denominator doesnt give us the numenator
    ??
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