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Integrate x^3/2 divided by expression - using partial fractions perhaps

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi. My first post!
    I'm trying to solve for where a is a constant:

    ∫ (x/a)1/2*(x/(x-a)) dx

    2. Relevant equations
    See above

    3. The attempt at a solution
    I've tried integration by parts by setting u=(x/a)1/2 but I end up having to solve ∫ (x/a)1/2ln(x-a) - which I can't solve.
    I've tried switching them around u=x/(x-a) and end up having to solve ∫x3/2/(x-a)2 - which I cant solve either.
    I've thought of using partial fractions but run into x3/2/(x-a)2 which I can't do either.

    Any help gratefully received.
  2. jcsd
  3. Apr 20, 2012 #2
    The change of variable u=(x/a)1/2 is very much solvable. Try it again.
  4. Apr 21, 2012 #3
    So obvious. I was looking for something really complicated.
    Let u = (x/a)1/2
    So x=au2, dx/du=2au

    First simplify:
    ∫(x/a)1/2(x/(x-a)) dx
    ∫(x/a)1/2( 1 + a/(x-a) ) dx

    Substitute by u:
    ∫u( 1 + a/(au2-a) ) 2au du
    2a∫u2( 1 + 1/(u2-1) ) du
    2/3au3 + 2a∫u2/(u2-1) du
    2/3au3 + 2a∫(1 + 1/(u2-1)) du
    2/3au3 + 2au + aln( (u-1)/(u+1) )

    Finally put back x and get required anaswer:
    2/3(x3/a)1/2 + 2(ax)1/2 + aln( ((x/a)1/2-1)/((x/a)1/2+1) )

    Thanks, for the hint which lead to the solution (was on it for weeks LOL)
    Last edited: Apr 21, 2012
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