Results 1 to 5 of 5

Math Help - Integral with complex denominator

  1. #1
    Member
    Joined
    Jul 2005
    Posts
    187

    Integral with complex denominator

    \int\frac{8x+3}{4x^2+12x+11}dx

    I am not able to solve it using partial fractions because denominator doesn't have real values. What should I do ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by totalnewbie View Post
    \int\frac{8x+3}{4x^2+12x+11}dx

    I am not able to solve it using partial fractions because denominator doesn't have real values. What should I do ?
    In that case you write it as,

    \frac{8x+3}{4x^2+12x+11}=\frac{Ax+B}{4x^2+12x+11}+  \frac{Cx+D}{4x^2+12x+11}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2005
    Posts
    187
    Quote Originally Posted by ThePerfectHacker View Post
    In that case you write it as,

    \frac{8x+3}{4x^2+12x+11}=\frac{Ax+B}{4x^2+12x+11}+  \frac{Cx+D}{4x^2+12x+11}
    Does it work out ?
    \int\frac{8x+3}{4x^2+12x+11}dx=\frac{1}{2}\int\fra  c{8x+12}{4x^2+12x+11}dx-\int\frac{8}{4x^2+12x+11}dx
    Is my approach right ? I am not sure if constant \frac{1}{2} is right.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by totalnewbie View Post
    Does it work out ?
    \int\frac{8x+3}{4x^2+12x+11}dx=\frac{1}{2}\int\fra  c{8x+12}{4x^2+12x+11}dx-\int\frac{8}{4x^2+12x+11}dx
    Is my approach right ? I am not sure if constant \frac{1}{2} is right.
    I do not think so, with or without the konstant.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    614
    Hello, totalnewbie!

    \int\frac{8x+3}{4x^2+12x+11}\,dx

    We have: . \int\frac{8x + 12 - 9}{4x^2+12x+11}\,dx \:=\:\int\frac{8x + 12}{4x^2+12x+11}\,dx \:- 9\int\frac{dx}{4x^2+12x+11}

    The first integral is of the form: . \int\frac{du}{u}


    For the second integral, complete the square:
    . . 4\left(x^2 + 3x + \frac{11}{4}\right) \;= \;4\left(x^2 + 3x + \frac{9}{4} + \frac{11}{4} - \frac{9}{4}\right) \;=\;4\left(\left[x+\frac{3}{2}\right]^2 + \frac{1}{2}\right)

    The integral becomes: . -\frac{9}{4}\int\frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}

    Trig Substitution: .Let x + \frac{3}{2}\:=\:\frac{1}{\sqrt{2}}\tan\theta . . . etc.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Definite integral with cosine at denominator
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 12th 2012, 08:18 AM
  2. [SOLVED] Complex Integral in Complex Analysis
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: August 21st 2011, 09:46 PM
  3. Replies: 12
    Last Post: February 15th 2011, 04:12 PM
  4. Can't factor the denominator in this integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 27th 2009, 05:44 PM
  5. limit with integral in denominator
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 22nd 2009, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum