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.
 A: Using substitution (as hinted in the comments), we get $\int_{0}^{1} x/2\cdot \log(x^2+1) - x/2\cdot \log(x^2+2)\; dx$. Let $u = x^2+1$ and $v = x^2 + 2$. Then we have:
$$\int_{1}^{2} du/4\cdot \log(u) - \int_{2}^{3} dv/4\cdot \log(v) \\
= 1/4\cdot [(\log(4/1) - 1)-(\log(27/4)-1)] \\
= 1/4\cdot \log(16/27) = -1/4\cdot \log(27/16).$$
Note: Perhaps the antiderivative of $\log(x)$ is not "obvious"; at least not so when you do it the first time. One could do this by integration by parts. Or consider $f(x) = x\log(x)$. Then $f'(x) = \log(x) + 1$ and so $[f(x)-x]' = f'(x) - 1 = \log(x)$ as desired. 
A: Let $u=1+x^2+y^2\;\Rightarrow\;du=2y\ dy$, then
$$
\begin{align}
\require{cancel}
\int_{y=-1}^0\frac{xy}{1+x^2+y^2}\ dy&=\int_{y=-1}^0\frac{x\cancel y}{u}\cdot \frac{du}{2\cancel y}\\
&=\frac x2\int_{y=-1}^0\frac1u\ du\\
&=\left.\frac x2\ln u\ \right|_{y=-1}^0\\
&=\left.\frac x2\ln (1+x^2+y^2)\ \right|_{y=-1}^0\\
&=\frac x2\left(\ln(1+x^2)-\ln(2+x^2)\right).\\
\end{align}
$$
The last integrals can be solved term by term by using substitution $u=1+x^2$ and $v=2+x^2$, then use IBP. The integration of primitive $\ln y$ can easy be found all over internet.
A: $\int_{0}^{1}\int_{-1}^{0}\dfrac{xy}{1 + x^2 + y^2}dydx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{2ydy}{1+x^2+y^2}dx = \int_{0}^{1}\dfrac{x}{2}\int_{-1}^{0}\dfrac{d(1 + x^2 + y^2)}{1 + x^2 + y^2}dx$. 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,
$-\dfrac{1}{4}\int_{1}^{2}[\ln(1 + u) - \ln u]du = -\dfrac{1}{4}\biggl[(1 + u)\ln(1 + u) - 1 - u - u\ln u + u\biggr]_{1}^{2} = -\dfrac{1}{4}[3\ln 3 - 2\ln 2 - 1 - 2\ln 2 + 1] = -\dfrac{3}{4}\ln 3 + \ln 2$
A: $$\int_0^1\int_{-1}^0\frac{xy}{x^2+y^2+1}dydx = \int_0^1x\int_{-1}^0y(x^2+y^2+1)^{-1}dydx\\= \int_0^1\frac{x}{2}\int_{-1}^0(x^2+y^2+1)^{-1}(2y)dydx \\= \int_0^1\frac{x}{2}\log(x^2+y^2+1)_{y=-1}^{0} dx \\= \int_0^1\frac{x}{2}(\log(x^2+1)-\log(x^2+2))dx \\= \frac{1}{4}\int_0^1\log(x^2+1)(2x)dx-\frac{1}{4}\int_0^1\log(x^2+2)(2x)dx\\=\frac{1}{4}(-x^2-1+(x^2+1)\log(x^2+1))-\frac{1}{4}(-x^2-2+(x^2+2)\log(x^2+2))_{x=0}^1\\=\frac{1}{4}(4 \log(2)-3\log(3))$$
