You can judge this one by its cover.

The integrand is $F(x,y) \; = \; 1 + \frac{xy}{1 + F(x,y)} \;= \; \sqrt{1 + xy}$ and so the integral is $\begin{aligned} I & = \; \int_0^1 \int_0^1 F(x,y)\,dx\,dy \; = \; \int_0^1 \,du \int_u^1 \frac{\sqrt{1+u}}{v}\,dv \; = \; -\int_0^1 \ln u \sqrt{1+u}\,du \\ & = \; \tfrac23\int_0^1 \frac{(1+u)^{\frac32} - 1}{u}\,du \end{aligned}$ using first the change of variables $u= xy\,,\, v = x$ , and then integrating by parts. Now the substitution $z = \sqrt{1+u}$ gives $\begin{aligned} I & = \; \tfrac23\int_1^{\sqrt{2}} \frac{z^3 - 1}{z^2-1} \times 2z\,dz \; = \; \tfrac43\int_1^{\sqrt{2}}\frac{z^4-z}{z^2-1}\,dz \\ & = \; \tfrac43\int_1^{\sqrt{2}} \left(z^2 + 1 - \frac{1}{z+1}\right)\,dz \; = \; \tfrac43\Big[\tfrac13z^2 + z - \ln(z+1)\Big]_1^{\sqrt{2}} \\ & = \; \tfrac43\Big[-\tfrac43 + \tfrac53\sqrt{2} + \ln2 - \ln(1 + \sqrt{2})\Big] \\ & = \; \tfrac43\Big[-\tfrac43 + \tfrac53\sqrt{2} + \ln\Big(\frac{4}{2\big(1 + \sqrt{\frac{4}{2}}\big)}\Big)\Big] \end{aligned}$ making the answer $4+3+5=\boxed{12}$ .