Mathematica gives

( a ArcTan[(b - a c)/(a Sqrt[1 - c^2])] +

b c ArcTan[(a - b c)/(b Sqrt[1 - c^2])] +

(a + b c) ArcSin[c])/Sqrt[1 - c^2] +

b/2 Log[(a^2 - 2 b c a + b^2)/b^2]

For example, with a -> 1, b -> 2, and c -> 1/2, one obtains

4 Pi/(3 Sqrt[3]) - Log[4/3]

which to 20 decimals is 2.1307170...

Doing the integral numerically,

NIntegrate[y/(x^2 + y^2 - 2 c y x) /. c -> 1/2, {x, 0, 2}, {y, 0, 1}]

2.1307170...

Cheers,

Paul

--

Paul Abbott Phone: +61 8 6488 2734

School of Physics, M013 Fax: +61 8 6488 1014

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