First, write the integral as a sum:
#int_{-3}^{1}(1+sqrt(9-x^2))dx=int_{-3}^{1}1dx+int_{-3}^{1}sqrt(9-x^2)dx#.
The area under the graph of #y=1# for #-3<=x<=1# is the area of a rectangle: #base * height=4*1=4#. Hence, #int_{-3}^{1}1dx=4#.
Next, write #int_{-3}^{1}sqrt(9-x^2)dx=int_{-3}^{0}sqrt(9-x^2)dx+int_{0}^{1}sqrt(9-x^2)dx#.
The graph of #y=sqrt(9-x^2)# is a semicircle of radius 3 centered at the origin. For #-3<=x<=0#, we get a quarter circle of area #1/4 pi*3^2=(9pi)/4#.
Hence, #int_{-3}^{0}sqrt(9-x^2)dx=(9pi)/4#.
The remaining region to calculate the area is the area under the graph of #y=sqrt(9-x^2)# for #0<=x<=1#. This region can be split into a triangle and a sector of the circle. See the figure below.
The triangle has #base=1# and #height=sqrt(9-1^2)=sqrt(8)=2sqrt(2)#. Therefore, its area is #1/2 * 1 * 2sqrt(2)=sqrt(2)#.
The angle for the sector of the circle is congruent to the upper angle for the triangle, which can be written as #theta=arcsin(1/3)# since the hypotenuse of the triangle has length 3. The area of the sector is therefore #1/2*r^2*theta=1/2*3^2*arcsin(1/3)=9/2 arcsin(1/3)#.
Therefore, #int_{0}^{1}sqrt(9-x^2)dx=sqrt(2)+9/2 arcsin(1/3)#.
Putting these things all together gives the final answer of #4+sqrt(2)+(9pi)/4+9/2 arcsin(1/3) approx 14.0121#
