<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] $\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{2ydy}{1+x^2+y^2}dx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{d(1 + x^2 + y^2)}{1 + x^2 + y^2}dx$. Thus,
$\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\ln(1 + x^2 + y^2)\mid_{-1}^{0}dx = -\int_{0}^{1}\dfrac{x}{2}\ln\biggl(1 + \dfrac{1}{1 + x^2}\biggr)dx$
If $u = 1 + x^2$, then $du = 2xdx$ and
$-\int_{0}^{1}\dfrac{x}{2}\ln\biggl(1 + \dfrac{1}{1 + x^2}\biggr)dx = -\dfrac{1}{4}\int_{1}^{2}\ln(1 + 1/u)du = -\dfrac{1}{4}\int_{1}^{2}[\ln(1 + u) - \ln u]du$
But, $\int \ln x dx = x\ln x - x + C$. Therefore,
$-\dfrac{1}{4}\int_{1}^{2}[\ln(1 + u) - \ln u]du = -\dfrac{1}{4}\biggl[(1 + u)\ln(1 + u) - 1 - u - u\ln u + u\biggr]_{1}^{2} = -\dfrac{1}{4}[3\ln 3 - 2\ln 2 - 1 - 2\ln 2 + 1] = -\dfrac{3}{4}\ln 3 + \ln 2$
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