Answer
$$\ln \left( {\frac{5}{3}} \right)$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\frac{x}{{{{\left( {1 + xy} \right)}^2}}}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 4,\,\,\,\,1 \leqslant y \leqslant 2} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^4 {\int_1^2 {\frac{x}{{{{\left( {1 + xy} \right)}^2}}}} } dydx \cr
& = \int_0^4 {\left[ {\int_1^2 {\frac{x}{{{{\left( {1 + xy} \right)}^2}}}} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \int_1^2 {\frac{x}{{{{\left( {1 + xy} \right)}^2}}}} dy \cr
& = \int_1^2 {{{\left( {1 + xy} \right)}^{ - 2}}\left( x \right)} dy \cr
& {\text{integrating by the power rule}} \cr
& = \left( {\frac{{{{\left( {1 + xy} \right)}^{ - 1}}}}{{ - 1}}} \right)_1^2 \cr
& = - \left( {\frac{1}{{1 + xy}}} \right)_1^2 \cr
& {\text{evaluating the limits for the variable }}y \cr
& = - \frac{1}{{1 + 2x}} + \frac{1}{{1 + x}} \cr
& \cr
& \int_0^4 {\left[ {\int_1^2 {\frac{x}{{{{\left( {1 + xy} \right)}^2}}}} dy} \right]} dx = \int_0^4 {\left( {\frac{1}{{1 + x}} - \frac{1}{{1 + 2x}}} \right)} dx \cr
& {\text{integrating}}{\text{, }} \cr
& = \left( {\ln \left| {1 + x} \right| - \frac{1}{2}\ln \left| {1 + 2x} \right|} \right)_0^4 \cr
& {\text{evaluate}} \cr
& = \ln \left| {1 + 4} \right| - \frac{1}{2}\ln \left| {1 + 2\left( 4 \right)} \right| - \ln \left| {1 + 0} \right| + \frac{1}{2}\ln \left| {1 + 2\left( 0 \right)} \right| \cr
& = \ln 5 - \frac{1}{2}\ln 9 \cr
& = \ln 5 - \ln 3 \cr
& = \ln \left( {\frac{5}{3}} \right) \cr} $$
