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Integral of

  1. Nov 19, 2009 #1
    hi,

    can anyone tell me how the following will be integrated:
    (where all letters except 'x' are constants)

    x/(ax^3 + bx -c )

    i tried to simplify the integral using partial fractions, but ended up with:

    p/(x-q) + (rx +s)/(tx^2 + ux + v)
    obviously, the first term is trivial, but how to integrate the second term. (the quadratic expression in the denominator does not have real roots)
     
  2. jcsd
  3. Nov 19, 2009 #2
    (rx +s)/(tx^2 + ux + v)

    If the numerator is a constant times the derivative of the denominator, great, make a substitution.

    Otherwise, complete the square in the denominator ... surely every calculus textbook has this method?
     
  4. Nov 19, 2009 #3

    HallsofIvy

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    If a quadratic expression such as [itex]tx^2+ ux+ v[/itex] cannot be factored, then you can complete the square to get something like [itex]t(x-a)^2+ b[/itex] where b is positive.
    Then [itex](rx+ x)/(t(x-a)^2+ b)[/itex] can be written as [itex](r(x-a)+ ra+ a)/(t(x-a)^2+ b)[/itex].
    The first term, [itex]r(x-a)dx/(t(x-a)^2+ b)[/itex] can be integrated with the substitution u= (x-a)^2 so that (1/2)du= (x-a)dt and the integrand becomes (r/(2t))du/u. The second term, (ra+a)dx/(t(x-a)^2+ b)= 1/(b(ra+a)) dx/((t/b)(x-a)^2+ 1) can be integrated with the substitution [itex]u= \sqrt{t}{b}(x-a)[/itex] and the fact that [itex]\int du/(u^2+1)= arctan(u)+ C[/itex].
     
  5. Nov 19, 2009 #4
    thanks guys!

    definite help! thanks!
     
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