<s>[INST] Double integral for $\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$ I'm trying to evaluate this
$$\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$$
tried substition
$$ u = {(x^2+y^2+1)}^{-1} \ \ du = \ln {(x^2+y^2+1)}$$ 
but du is not found in the given equation. I have the feeling that I should use arctan but I do not know how to apply that in two variables.
 [/INST] $$\int_0^1\int_{-1}^0\frac{xy}{x^2+y^2+1}dydx = \int_0^1x\int_{-1}^0y(x^2+y^2+1)^{-1}dydx\\= \int_0^1\frac{x}{2}\int_{-1}^0(x^2+y^2+1)^{-1}(2y)dydx \\= \int_0^1\frac{x}{2}\log(x^2+y^2+1)_{y=-1}^{0} dx \\= \int_0^1\frac{x}{2}(\log(x^2+1)-\log(x^2+2))dx \\= \frac{1}{4}\int_0^1\log(x^2+1)(2x)dx-\frac{1}{4}\int_0^1\log(x^2+2)(2x)dx\\=\frac{1}{4}(-x^2-1+(x^2+1)\log(x^2+1))-\frac{1}{4}(-x^2-2+(x^2+2)\log(x^2+2))_{x=0}^1\\=\frac{1}{4}(4 \log(2)-3\log(3))$$
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