Answer
$$\frac{{{e^3} - e - {e^{ - 1}} + {e^{ - 3}}}}{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\int_{y - 1}^{y + 1} {{e^{x + y}}} dxdy} \cr
& {\text{Recall that }}{e^a}{e^b} = {e^{a + b}} \cr
& \int_{ - 1}^1 {\int_{y - 1}^{y + 1} {{e^x}{e^y}} dxdy} \cr
& \int_{ - 1}^1 {{e^y}\int_{y - 1}^{y + 1} {{e^x}} dxdy} \cr
& \int_{ - 1}^1 {{e^y}\left[ {\int_{y - 1}^{y + 1} {{e^x}} dx} \right]dy} \cr
& {\text{Evaluate the inner integral}} \cr
& \int_{y - 1}^{y + 1} {{e^x}} dx = \left[ {{e^x}} \right]_{x = y - 1}^{x = y + 1} \cr
& = {e^{y + 1}} - {e^{y - 1}} \cr
& {\text{Therefore,}} \cr
& \int_{ - 1}^1 {{e^y}\left[ {\int_{y - 1}^{y + 1} {{e^x}} dx} \right]dy} \cr
& = \int_{ - 1}^1 {{e^y}\left( {{e^{y + 1}} - {e^{y - 1}}} \right)dy} \cr
& = \int_{ - 1}^1 {\left( {{e^{2y + 1}} - {e^{2y - 1}}} \right)dy} \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{1}{2}{e^{2y + 1}} - \frac{1}{2}{e^{2y - 1}}} \right]_{ - 1}^1 \cr
& = \left[ {\left( {\frac{1}{2}{e^{2\left( 1 \right) + 1}} - \frac{1}{2}{e^{2\left( 1 \right) - 1}}} \right) - \left( {\frac{1}{2}{e^{2\left( { - 1} \right) + 1}} - \frac{1}{2}{e^{2\left( { - 1} \right) - 1}}} \right)} \right] \cr
& = \left[ {\frac{1}{2}{e^3} - \frac{1}{2}{e^1}} \right] - \left[ {\frac{1}{2}{e^{ - 1}} - \frac{1}{2}{e^{ - 3}}} \right] \cr
& = \frac{1}{2}{e^3} - \frac{1}{2}e - \frac{1}{2}{e^{ - 1}} + \frac{1}{2}{e^{ - 3}} \cr
& = \frac{{{e^3} - e - {e^{ - 1}} + {e^{ - 3}}}}{2} \cr} $$
