## difficult integral

(x/(x-a))^0.5 where "a" is a start distance of 10^-3 and the final distance needs to be 10^-2

It looks simple but its not.

Integrate[(x/(x - a))^0.5, x] ==
(0.*(x/(-a + x))^0.5*(-a + x)^0.5)/x^0.5 + (2.*(x/(-a + x))^0.5*(-a + x)^0.5*(-1.*a + x)^0.5* Hypergeometric2F1[0.5, -0.5, 1.5, 1. - (1.*x)/a])/ (0. + x/a)^0.5

Which is way way over my head. Is there a simpler solution?

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 Quote by Per Oni Wolfram integrator gave this answer: Integrate[(x/(x - a))^0.5, x] == (0.*(x/(-a + x))^0.5*(-a + x)^0.5)/x^0.5 + (2.*(x/(-a + x))^0.5*(-a + x)^0.5*(-1.*a + x)^0.5* Hypergeometric2F1[0.5, -0.5, 1.5, 1. - (1.*x)/a])/ (0. + x/a)^0.5 Which is way way over my head. Is there a simpler solution?
Tell Wolfram again, but this time use 1/2 and not 0.5 ... this tells Wolfram that the exponent is an exact number, and not just a decimal approximation to some number. If the exponent is very close to 1/2, but perhaps not equal to 1/2, then the answer will come out as an 2F1 as shown. But if the exponent is exactly 1/2, then you can get an answer in logarithms.

## difficult integral

 Quote by g_edgar Tell Wolfram again, but this time use 1/2 and not 0.5 ... this tells Wolfram that the exponent is an exact number, and not just a decimal approximation to some number. If the exponent is very close to 1/2, but perhaps not equal to 1/2, then the answer will come out as an 2F1 as shown. But if the exponent is exactly 1/2, then you can get an answer in logarithms.
Thanks a lot. Using 1/2 gave me a sensible answer.