Results 1 to 4 of 4

Math Help - Integrate by partial fractions

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    81

    Integrate by partial fractions

    \displaystyle \int \frac{6 x^3+4 x^2 - 3 x - 5}{x^2-1}\, dx

    What am I doing wrong here?

    First, I long divide & get:

    <br />
\int 6x+4+\frac{3x-1}{x^2-1}dx<br />

    \int 6x+4+\frac{3x-1}{(x+1)(x-1)}dx= \int 6x+4 +\frac{A}{x+1}+\frac{B}{x-1} dx

    <br />
A(x-1)+B(x+1)=3x-1<br />

    I figure that A = 1, B = 2

    <br />
\int 6x+4 + \frac{2}{x+1} + \frac{1}{x-1} dx<br />


    This should integrate to...

    <br />
3x^2+4x+2ln(x+1) + ln(x-1) + C<br />

    this answer appears to be wrong...
    Where did I go wrong?

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    the answer is right!

    Check it here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Vamz View Post
    \displaystyle \int \frac{6 x^3+4 x^2 - 3 x - 5}{x^2-1}\, dx

    What am I doing wrong here?

    First, I long divide & get:

    <br />
\int 6x+4+\frac{3x-1}{x^2-1}dx<br />

    \int 6x+4+\frac{3x-1}{(x+1)(x-1)}dx= \int 6x+4 +\frac{A}{x+1}+\frac{B}{x-1} dx

    <br />
A(x-1)+B(x+1)=3x-1<br />

    I figure that A = 1, B = 2

    <br />
\int 6x+4 + \frac{2}{x+1} + \frac{1}{x-1} dx<br />


    This should integrate to...

    <br />
3x^2+4x+2ln(x+1) + ln(x-1) + C<br />

    this answer appears to be wrong...
    Where did I go wrong?

    Thank you!
    I think you'll find that A = 2 and B = 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Vamz View Post
    <br />
3x^2+4x+2ln(x+1) + ln(x-1) + C<br />
    I must be getting picky in my old age....

    The solution is
    3x^2+4x+2ln|x+1| + ln|x-1| + C

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 28th 2010, 09:53 AM
  2. Partial fractions integrate
    Posted in the Calculus Forum
    Replies: 15
    Last Post: March 27th 2010, 02:41 PM
  3. Integrate problem with partial fractions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 27th 2009, 07:41 AM
  4. Integrate using partial fractions
    Posted in the Calculus Forum
    Replies: 14
    Last Post: June 27th 2008, 08:30 PM
  5. Integrate with Improper Fractions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 9th 2007, 08:05 AM

Search Tags


/mathhelpforum @mathhelpforum