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Evaluating an integral involving a fraction

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data

    I want to integrate:

    [itex]\int (\frac{x-a}{b-x})^{c/d}dx[/itex]

    2. Relevant equations

    given:
    c<d

    3. The attempt at a solution

    I have no idea which integration method to use. i tried to make the integrand look easier by rewriting it as:

    [itex](1+\frac{b-a}{x-b})^{c/d}[/itex]

    but this still doesnt help me.
    any ideas will be very much appreciated. thank you.
     
  2. jcsd
  3. Jul 17, 2011 #2

    hunt_mat

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    Have you tried the obvious one?
    [tex]
    u=\frac{x-a}{b-x}
    [/tex]
     
  4. Jul 17, 2011 #3
    after using the substitution:

    [itex]u=\frac{x-a}{b-x}[/itex],

    we find the derivative of u with respect to x so that:

    [itex]du=\frac{b-a}{(b-x)^2}dx[/itex]
    (using the quotient rule for differentiation),

    then, substituting this into the original integral, we have:

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

    and from here, i can't seem to think of another substitution.
    any ideas?
    (thank you in advance)
     
  5. Jul 17, 2011 #4
    Is c supposed to represent the dummy constant when evaluating an indefinite integral, or is it just a generic variable here like a and b?
     
  6. Jul 17, 2011 #5
    it is a generic variable like a and b.
     
  7. Jul 17, 2011 #6

    hunt_mat

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    Hmm, that ends up quite messy and not very helpful in the slightest (but you did it correctly), you did make another (very nice) observation that you integrand becomes:
    [tex]
    \left(1+\frac{b-a}{b-x}\right)^{\frac{c}{d}}
    [/tex]
    Now I would use the following substitution:
    [tex]
    u=\frac{1}{b-x}
    [/tex]
     
    Last edited: Jul 17, 2011
  8. Jul 17, 2011 #7
    I tried this substitution u=1/(x-b),
    so du=ln(x-b)dx

    so, substituting this into the integral, we have:

    [itex]\int\frac{(1+(b-a)u)^{c/d}}{-ln(u)}du[/itex]

    and... also, i am stuck on what to chose for the next substitution.
     
  9. Jul 17, 2011 #8

    hunt_mat

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    That is not what you get, you integrated instead of differentiated, what is du?
     
  10. Jul 17, 2011 #9
    my mistake.

    du=-1/(x-b)^2 dx

    so it should be:

    [itex]\int\frac{(1+(b-a)u)^{c/d}}{u^2}du[/itex]
     
  11. Jul 17, 2011 #10

    hunt_mat

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    Any thoughts now?
     
  12. Jul 18, 2011 #11
    i can't think of a further useful substitution, because we have the power of c/d.
     
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