Answer
$$3 \sqrt {11}+2 \sinh^{-1} (\dfrac{3}{\sqrt 2})$$
Work Step by Step
The surface area of the part $z=f(x,y)$ can be computed as: $A(S)=\iint_{D} \sqrt {1+(z_x)^2+(z_y)^2} dx \ dy$
and, $\iint_{D} dA$ represents the projection of the surface on the $xy$-plane.
The area of the given surface is:
$A(S)=\iint_{D} \sqrt{(2x+1)^{2}+(1)^{2}+1} d A \\ =\iint_{D} \sqrt{4x^2+4x+1+1+1} d A \\ =\iint_{D} \sqrt{4x^2+4x+3} d A $
Since, $-2 \leq x \leq 1$ and $-1 \leq y \leq 1$
Now, we will use a calculator to find the area of the given surface.
$$A(S)=\iint_{D} \sqrt{4x^2+4x+3} d A \\= \int_{-1}^{1} \int_{-2}^{1} \sqrt{4x^2+4x+3} \space dx \space dy \\=3 \sqrt {11}+2 \sinh^{-1} (\dfrac{3}{\sqrt 2})$$
