Solving Difficult Integral: x/(x-a)^0.5 from 10^-3 to 10^-2

[Per Oni]

Please help required with this 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.
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?

 

[Gib Z]

Assuming the integral you want to solve is \int \sqrt{ \frac{x}{x-a}} dx, make the substitution x= a \sec^2 \theta.

 

[g_edgar]

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.

 

[Per Oni]

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.