Results 1 to 6 of 6

Math Help - How would you integrate this?

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    105

    How would you integrate this?

    5x
    (x-1)((x^2)+4)

    fractions?
    5x = A + B
    (x-1) ?

    And ((x^2)+4) is 1/2arctan(x/2)

    the answer key says
    ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)

    Can anyone help? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    \frac{5x}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+4} (since x^2 + 4 is an irreducible quadratic, indicating your numerator is in the form of Bx + C)

    See what you can do with this and come back with any questions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by khuezy View Post
    5x
    (x-1)((x^2)+4)

    fractions?
    5x = A + B
    (x-1) ?

    And ((x^2)+4) is 1/2arctan(x/2)

    the answer key says
    ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)

    Can anyone help? Thanks.
    You can use partial fractions with this setup

    \frac{5x}{(x^2+4)(x-1)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+4}

    Or observe that

    \frac{5x}{(x-1)(x^2+4)}=\frac{5x-x^2-4+x^2+4}{(x-1)(x^2+4)}=\frac{-(x^2-5x+4)}{(x-1)(x^+2)}+\frac{x^2+4}{(x-1)(x^2+4)}=

    \frac{-(x-4)(x-1)}{(x^2+4)(x-1)}+\frac{x^2+4}{(x-1)(x^2+4)}=\frac{4-x}{x^2+4}+\frac{1}{x-1}=\frac{4}{x^2+4}-\frac{x}{x^2+4}+\frac{1}{x-1}

    Now we have

    \int \left( \frac{4}{x^2+4}-\frac{x}{x^2+4}+\frac{1}{x-1} \right)dx

    The first is in the form of the arctangent or if you want to do it by hand let x=2tan(u) the 2nd is a u sub on the denominator u=x^2+4 and the last one is a u sub on the denominator u=x-1.

    I hope this helps.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2008
    Posts
    105

    help

    I got
    5x = (A+B)x^2 + 4A - (B+C)x - C

    I'm confused on what to do next.

    Equating coefficients?
    A+B = 0 4A=5 B+C=0 C=0????????????????

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Here's another way using trig.

    \frac{x-1}{x^{2}+4}dx

    Let x=2tan(t), \;\ dx=2sec^{2}(t)dt

    Making the subs we get a menacing looking thing, but it simplifies down fairly nice.

    Making the subs we get:

    \frac{2tan(t)-1}{4(tan^{2}(t)+1)}\cdot{2sec^{2}(t)}dt

    This, believe it or not, whittles down to

    tan(t)-\frac{1}{2}

    \int{tan(t)}dt-\frac{1}{2}\int{dt}

    Then, after integrating, resub in t=tan^{-1}(\frac{x}{2})

    You may wish to stick with the others methods. I just threw it out there.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by khuezy View Post
    5x
    (x-1)((x^2)+4)

    fractions?
    5x = A + B
    (x-1) ?

    And ((x^2)+4) is 1/2arctan(x/2)

    the answer key says
    ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)

    Can anyone help? Thanks.
    \frac{5x}{(x^2+4)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}

    multiplying through by the common denominator we get

    5x=A(x^2+4)+(Bx+C)(x-1)

    Now letting x=1 we get

    5(1)=A(1^2+4)+(B+C)(0)\Rightarrow{A=1}

    So now for the second factor we need to expand, but dont forget A=1

    So we have

    5x=x^2+4+Bx^2-Bx+Cx-C

    So we see that B+1=0\Rightarrow{B=-1}

    and also seeing 4-C=0\Rightarrow{C=4}

    So finally we can see that

    \frac{5x}{(x^2+4)(x-1)}=\frac{1}{x-1}+\frac{4-x}{x^2+4}
    and we see that
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integrate 3 e^x
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 9th 2010, 01:16 PM
  2. how to integrate this ??
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 9th 2009, 08:14 PM
  3. Integrate
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 28th 2009, 07:54 PM
  4. How to integrate x^x?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2009, 07:51 PM
  5. Integrate
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 15th 2009, 03:50 PM

Search Tags


/mathhelpforum @mathhelpforum