Results 1 to 9 of 9

Math Help - Partial Fractions

  1. #1
    Junior Member symstar's Avatar
    Joined
    Jul 2007
    From
    Berkeley, CA
    Posts
    57

    Partial Fractions

    I need to do this without using complex numbers.

    \int \frac{10}{(x-1)(x^2+9)}dx

    Working with the fraction:
    \frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}
    10=A(x^2+9)+(Bx+C)(x-1)
    10=(A+B)x^2+(-B+C)x+(9A-C)
    C=-1; B=-1; A=1

    Back to the integral:
    \int\frac{1}{x-1}dx-\int\frac{x+1}{x^2+9}dx

    Assuming I made it this far correctly, I'm not quite sure what to do with the second integral... obviously the first is \ln\left|x-1\right|
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,115
    Thanks
    992
    \frac{x+1}{x^2 + 9} = \frac{x}{x^2 + 9} + \frac{1}{x^2 + 9}<br />

    integral of the first fraction will be \ln\sqrt{x^2+9}

    integral of the second will be \frac{1}{3}\arctan \left(\frac{x}{3}\right)
    Last edited by skeeter; September 14th 2008 at 06:23 PM. Reason: fix constant multiplier
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by symstar View Post
    I need to do this without using complex numbers.

    \int \frac{10}{(x-1)(x^2+9)}dx

    Working with the fraction:
    \frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}
    10=A(x^2+9)+(Bx+C)(x-1)
    10=(A+B)x^2+(-B+C)x+(9A-C)
    C=-1; B=-1; A=1

    Back to the integral:
    \int\frac{1}{x-1}dx-\int\frac{x+1}{x^2+9}dx

    Assuming I made it this far correctly, I'm not quite sure what to do with the second integral... obviously the first is \ln\left|x-1\right|
    Note that \int\frac{x+1}{x^2+9}dx=\int\frac{x}{x^2+9}dx+\int  \frac{1}{x^2+9}dx

    The first one can be done using substitution, and then the last one has the form of arctangent!

    Try to take it from here.

    --Chris
    Follow Math Help Forum on Facebook and Google+

  4. #4
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Basically split up your fraction into two:
    \int \frac{x+1}{x^2 + 9} \ dx \ = \ \int \frac{x}{x^2 + 9} \ dx + \int \frac{1}{x^2+9} \ dx
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    I didn't check your work but you can use a trig substituion to solve the integral you are having difficulty with.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member symstar's Avatar
    Joined
    Jul 2007
    From
    Berkeley, CA
    Posts
    57
    Looking back through my text, I realized I breezed over that section a little bit too quickly. Splitting into 2 integrals was what I missed. Anyways, I got the right answer, thanks everyone for your help!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    without partial fractions:

    \frac {10}{(x - 1)(x^2 + 9)} = \frac {x^2 + 9 + 1 - x^2}{(x - 1)(x^2 + 9)}

    ............................ = \frac {x^2 + 9}{(x - 1)(x^2 + 9)} + \frac {(1 - x)(x + 1)}{(x - 1)(x^2 + 9)}

    ............................ = \frac 1{x - 1} - \frac {x + 1}{x^2 + 9}

    ............................ = \frac 1{x - 1} - \frac 1{x^2 + 9} - \frac x{x^2 + 9}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Haha, it seems you started to give short answers, that's good, keep it up!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Krizalid View Post
    Haha, it seems you started to give short answers, that's good, keep it up!
    yup! i was like "Krizalid would be so proud of me "

    by the way, congrats on the 2th post!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 28th 2010, 10:53 AM
  2. Partial Fractions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 4th 2010, 07:24 PM
  3. Partial Fractions
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: January 17th 2010, 03:20 AM
  4. Help on Partial Fractions
    Posted in the Calculus Forum
    Replies: 12
    Last Post: January 6th 2010, 04:00 AM
  5. partial fractions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 21st 2007, 05:49 AM

Search Tags


/mathhelpforum @mathhelpforum