Comment #9 on issue 2219 by [email protected]: Arbitrary constants in
indefinite integration
http://code.google.com/p/sympy/issues/detail?id=2219
Yes, regarding free_symbols, that will give you all of the unbound symbols:
h[2] >>> Integral(x+y,(x,1,2)).free_symbols
set([y])
h[3] >>> Integral(x+y,x).free_symbols
set([x, y])
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