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Difficult integral.

  1. Aug 17, 2014 #1

    [itex]\int \frac{x}{(x^2+1)(bx+1)}[/itex]

    Is the problem, which I am trying to work through. "b" is just a constant, it is very tough to integrate this indefinitely, any help will be greatly appreciated. Off the top of this,

    I tried with the u-substitution,

    let [itex]u = bx + 1[/itex] and [itex]x = (u-1)/b [/itex]
    [itex]x^2 = [(u-1)/b]^2[/itex]
    [itex] du = b[/itex]

    = [itex](b)\int \frac{\frac{u-1}{b}}{\frac{u((u-1)^2+b^2)}{b^2}} du[/itex]

    = [itex](b)\int \frac{b(u-1)}{u[(u-1)^2+b^2]}[/itex]

    This is where I get lost.. Any help? Thanks
  2. jcsd
  3. Aug 17, 2014 #2
    Have you tried partial fraction decomposition?
  4. Aug 17, 2014 #3
    Yes try to do partial fractions, it will work out.
  5. Aug 17, 2014 #4
    Yes, look below
  6. Aug 18, 2014 #5


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    Partial Fraction Decomposition

    Set [itex]\frac{x}{(x^2+1)(bx+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{bx+1}\Rightarrow x= (Ax+B) (bx+1)+C(x^2+1)\Rightarrow x=(Ab+C)x^2+(A+Bb)x+(B+C)[/itex]

    so that we have the equations (by equating coefficients of like powers of x): [itex]Ab+C=0,A+Bb=1,B+C=0\Rightarrow A=1/(b^2+1),B=b/(b^2+1),C=-b/(b^2+1)[/itex]

    hence [itex]\frac{x}{(x^2+1)(bx+1)}=\frac{1}{b^2+1}\left(\frac{x+b}{x^2+1}-\frac{b}{bx+1}\right)[/itex]
  7. Aug 18, 2014 #6

    Yes, I got it. I got the correct answer; this was part of a Putnam integral by the way.This thread can now be locked.
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