Answer
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \frac{y}{{1 + xy}}{\rm{d}}y{\rm{d}}x \simeq 0.386294$
Work Step by Step
We use Fubini's Theorem to change the order of integration and re-write the integral:
$I = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \frac{y}{{1 + xy}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 0}^1 \frac{y}{{1 + xy}}{\rm{d}}x} \right){\rm{d}}y$
We evaluate first the inner integral with respect to $x$.
$S\left( y \right) = \mathop \smallint \limits_{x = 0}^1 \frac{y}{{1 + xy}}{\rm{d}}x$
The antiderivative of $\frac{y}{{1 + xy}}$ with respect to $x$ is $\ln \left( {1 + xy} \right)$ because
$\frac{\partial }{{\partial x}}\left( {\ln \left( {1 + xy} \right)} \right) = \frac{y}{{1 + xy}}$
So,
$S\left( y \right) = \mathop \smallint \limits_{x = 0}^1 \frac{y}{{1 + xy}}{\rm{d}}x = \ln \left( {1 + xy} \right)|_{x = 0}^1$
$ = \ln \left( {1 + y} \right)$
Thus,
$I = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 0}^1 \frac{y}{{1 + xy}}{\rm{d}}x} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^1 \ln \left( {1 + y} \right){\rm{d}}y$
We need an antiderivative of $\ln \left( {1 + y} \right)$ as a function of $y$.
We can use $y\ln \left( {1 + y} \right) + \ln \left( {1 + y} \right) - y$ because
$\frac{d}{{dy}}\left( {y\ln \left( {1 + y} \right) + \ln \left( {1 + y} \right) - y} \right) = \ln \left( {1 + y} \right) + \frac{y}{{1 + y}} + \frac{1}{{1 + y}} - 1$
$ = \ln \left( {1 + y} \right) + \frac{{y + 1}}{{1 + y}} - 1$
$ = \ln \left( {1 + y} \right)$
So,
$I = \left( {y\ln \left( {1 + y} \right) + \ln \left( {1 + y} \right) - y} \right)|_0^1$
$ = \ln 2 + \ln 2 - 1$
$ = 2\ln 2 - 1$
Therefore, $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \frac{y}{{1 + xy}}{\rm{d}}y{\rm{d}}x = 2\ln 2 - 1 \simeq 0.386294$.
