Answer
$$\int_{0}^{1} \int_{0}^{1} \frac{x}{(x y+1)^{2}} d y d x=1-\ln2$$
Work Step by Step
Given $$\int_{0}^{1} \int_{0}^{1} \frac{x}{(x y+1)^{2}} d y d x$$
So, we have
\begin{align}
I&=\int_{0}^{1} \int_{0}^{1} \frac{x}{(x y+1)^{2}} d y d x\\
&=\int_{0}^{1} \int_{0}^{1} x (x y+1)^{-2 } d y d x\\
&=-\int_{0}^{1} (x y+1)^{-1 }|_{0}^{1} d x\\
&=-\int_{0}^{1} (x +1)^{-1 } d x+\int_{0}^{1} (0 +1)^{-1 } d x\\
&=-\int_{0}^{1} \frac{1}{(x +1)} d x+\int_{0}^{1} d x\\
&=(-\ln (x +1) +x)_{0}^{1}\\
&=(-\ln (1 +1) +1)- (-\ln (0 +1) +0)\\
&=1-\ln 2
\end{align}
