Answer
\[\frac{1}{4}\ln \left( \frac{8}{17} \right)\]
Work Step by Step
\[\begin{align}
& \int_{1}^{2}{\int_{1}^{4}{\frac{xy}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}dx}}dy \\
& \text{Evaluate inner integral }\int_{1}^{4}{\frac{xy}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}dx} \\
& =\frac{y}{2}\int_{1}^{4}{\frac{2x}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}dx} \\
& =\frac{y}{2}\left[ -\frac{1}{{{x}^{2}}+{{y}^{2}}} \right]_{1}^{4} \\
& =-\frac{y}{2}\left[ \frac{1}{{{\left( 4 \right)}^{2}}+{{y}^{2}}}-\frac{1}{{{\left( 1 \right)}^{2}}+{{y}^{2}}} \right] \\
& =-\frac{y}{2}\left[ \frac{1}{{{y}^{2}}+16}-\frac{1}{{{y}^{2}}+1} \right] \\
& \text{Then} \\
& \int_{1}^{2}{\int_{1}^{4}{\frac{xy}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}dx}}dy=-\frac{1}{2}\int_{1}^{2}{\left[ \frac{y}{{{y}^{2}}+16}-\frac{y}{{{y}^{2}}+1} \right]}dy \\
& =-\frac{1}{4}\left[ \ln \left( {{y}^{2}}+16 \right) \right]_{1}^{2}+\frac{1}{4}\left[ \ln \left( {{y}^{2}}+1 \right) \right]_{1}^{2} \\
& =-\frac{1}{4}\left[ \ln \left( 20 \right)-\ln \left( 17 \right) \right]_{1}^{2}+\frac{1}{4}\left[ \ln \left( 5 \right)-\ln \left( 2 \right) \right] \\
& =-\frac{1}{4}\ln \left( \frac{20}{17} \right)+\frac{1}{4}\ln \left( \frac{5}{2} \right) \\
& =\frac{1}{4}\ln \left( \frac{8}{17} \right) \\
\end{align}\]
