Answer
$$\frac{{29}}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_0^{1 + x} {\left( {3x + 2y} \right)} } dydx \cr
& = \int_0^1 {\left[ {\int_0^{1 + x} {\left( {3x + 2y} \right)} } \right]dy} dx \cr
& {\text{Integrate with respect to }}y \cr
& \int_0^{1 + x} {\left( {3x + 2y} \right)} = \left[ {3xy + {y^2}} \right]_0^{1 + x} \cr
& = \left[ {3x\left( {1 + x} \right) + {{\left( {1 + x} \right)}^2}} \right] - \left[ {3x\left( 0 \right) + {{\left( 0 \right)}^2}} \right] \cr
& = 3x + 3{x^2} + 1 + 2x + {x^2} \cr
& = 4{x^2} + 5x + 1 \cr
& = \int_0^1 {\left[ {\int_0^{1 + x} {\left( {3x + 2y} \right)} } \right]dy} dx = \int_0^1 {\left( {4{x^2} + 5x + 1} \right)} dx \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{4}{3}{x^3} + \frac{5}{2}{x^2} + x} \right]_0^1 \cr
& {\text{Simplifying}} \cr
& = \left[ {\frac{4}{3}{{\left( 1 \right)}^3} + \frac{5}{2}{{\left( 1 \right)}^2} + \left( 1 \right)} \right] - \left[ {\frac{4}{3}{{\left( 0 \right)}^3} + \frac{5}{2}{{\left( 0 \right)}^2} + \left( 0 \right)} \right] \cr
& = \frac{4}{3} + \frac{5}{2} + 1 \cr
& = \frac{{29}}{6} \cr} $$
