Answer
$$\ \iint \limits_{R} \frac{x}{1+x y} d A = 2 \ln(2) - 1 $$
Work Step by Step
Given$$\ \iint \limits_{R} \frac{x}{1+x y} d A, \quad R=[0,1] \times[0,1]$$
So, we get \begin{aligned} I&= \int_{0}^{1} \int_{0}^{1} \frac{x}{1+x y} dy \ dx \\
&=\int_{0}^{1} \left[\ln (1+x y)\right]_{0}^{1} dx \\
&=\int_{0}^{1} \left[\ln (1+x )\right] dx -\int_{0}^{1} \left[\ln (1)\right] dx \\
&=\int_{0}^{1} \left[\ln (1+x )\right] dx-0\\
&=\int_{0}^{1} \left[\ln (1+x )\right] dx\\
\end{aligned}
So, by the partition technique let $$u=\ln (1+x ) \Rightarrow du= \frac{1}{1+x }dx$$ $$dv= dx \Rightarrow v=x $$ So, we get
\begin{aligned}
I&=\int_{0}^{1} \left[\ln (1+x )\right] dx\\
&=uv|_{0}^{1}- \int_{0}^{1} vdu\\
&=x \ln(1+x)|_{0}^{1}- \int _{0}^{1}\frac{x}{1+x }dx\\
&= \ln(2)-0 - \int _{0}^{1}\frac{1+x-1}{1+x }dx\\
&= \ln(2) - \int _{0}^{1}\left[1-\frac{ 1}{1+x } \right]dx\\
&= \ln(2) - \left[x-\ln (1+x ) \right]_{0}^{1} \\
&= \ln(2) - \left[1-\ln (1+1 ) -0+\ln1\right] \\
&= \ln(2) - 1+\ln (2 ) \\
&= 2 \ln(2) - 1 \\
\end{aligned}
