Results 1 to 15 of 15

Math Help - root integral question..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    root integral question..

    <br />
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=<br />
\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}<br />
    i got read of one root but instead i got another one
    ??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    Quote Originally Posted by transgalactic View Post
    <br />
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=<br />
\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}<br />
    i got read of one root but instead i got another one
    ??
    If you let x = u^6 then

    \int \frac{6\,du}{u(u+1)(2u^2-u+1)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    when i substituted i didnt get your expression
    i got
    <br />
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)}<br />
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42

    Thumbs up

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

    2u^3+u^2+ 1 = (u+1)(2u^2-u+1)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i cant solve this last part:
    <br />
\frac{1}{3}\int\frac{-27u+2}{2u^2-u+1} <br />

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

    ??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by transgalactic View Post
    i cant solve this last part:
    <br />
\frac{1}{3}\int\frac{-27u+2}{2u^2-u+1} <br />

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

    ??
    You should get <br />
\int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1} <br />

    You have to split into 2 parts

    \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 v = \frac{4t}{\sqrt{7}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    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; February 20th 2009 at 04:19 AM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    \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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    its still does give the last part as i got it

    where did my coefficientcient equations differ yours??
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    did i get the coefficient equations ..

    <br />
\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\<br />
    <br />
A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\<br />
    <br />
u^3:\\<br />
    <br />
2A-2B+C=0\\<br />
    <br />
u^2:\\<br />
    <br />
A+B+C+D=0\\<br />
u:\\<br />
    <br />
-B+D=0\\<br />
    1:
    <br />
A=1<br />
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Air
    Air is offline
    Junior Member Air's Avatar
    Joined
    Aug 2008
    Posts
    47
    Awards
    1
    Quote Originally Posted by transgalactic View Post
    <br />
\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\<br />
    <br />
A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\<br />
    <br />
u^3:\\<br />
    <br />
2A-2B+C=0\\<br />
    <br />
u^2:\\<br />
    <br />
A+B+C+D=0\\<br />
u:\\<br />
    <br />
-B+D=0\\<br />
    1:
    <br />
A=1<br />
    The denominator factorises to u(1+2u^3+u^2) = u(u+1)(2u^2 - u +1).

    Thus, The fraction becomes \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. B is +ve, and not -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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Your equations are correct.

    There is a much easier way :

    \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

    \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
    1 = A

    Multiply both sides by u+1
    \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
    -\frac{1}{4} = - B
    Last edited by mr fantastic; February 21st 2009 at 04:20 AM. Reason: Required editing because it's been merged from another thread
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i tried again

    here is my full way till i got the coefficients
    where did i go wrong
    ??
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    C=-9 is OK
    then -3+2D=0 gives D=3/2 and not 2/3
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    <br />
\int \frac{-9u+\frac{3}{2}}{2u^2-u+1}<br />

    but then the derivative of the denominator doesnt give us the numenator
    ??
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. root integral question
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 25th 2011, 08:12 AM
  2. Replies: 3
    Last Post: March 30th 2011, 07:46 PM
  3. integral (square root)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 10th 2010, 07:49 PM
  4. Replies: 4
    Last Post: June 30th 2010, 04:14 PM
  5. Integral with Sq Root of u on bottom question
    Posted in the Calculus Forum
    Replies: 8
    Last Post: July 11th 2009, 05:28 PM

Search Tags


/mathhelpforum @mathhelpforum