Exactly what do you mean by "closed form solution"? Since a limits of integration are numbers, the integral is, of course, a single number, which is about as "closed form" as you can get! But I expect that you are asking about a method of finding that number, which is a different matter.
As jedishrfu said, seeing that x^2+ y^2, I would try polar coordinates. Of course, the fact that the region of integration is a rectangle rather than a disk complicates things!
As the "\theta" line sweeps up from the x-axis (\theta= 0) it first touches the rectangle at (2, 3) (slope 3/2), then at the point (2, 4) (slope 2), then (1, 3) (slope 3), and finally at (1, 4) (slope 4). That gives three intervals, \theta= arctan(3/2) to arctan(2), \theta= arctan(2) to arctan(3), and \theta= arctan(3) to arctan(4) on which the ray enters and leaves the rectangle through different edges.
For \theta= arctan(3/2) to arctan(2), the ray enters the rectangle through the bottom, y= r sin(\theta)= 3 (so r= 3/sin(\theta)= 3 csc(\theta)) and leaves through the right side, x= rcos(\theta)= 2 (so r= 2/cos(\theta)=2 sec(\theta). That integral will be
\int_{arctan(3/2)}^{arctan(2)}\int_{3csc(\theta)}^{2sec(\theta)}\sqrt{1+4r^2} rdrd\theta.
For \theta= arctan(2) to arctan(3), the ray enters the rectangle through the bottom and leaves through top, y= r sin(\theta)= 4 (so r= 4/sin(\theta)= 4 csc(\theta)). That integral will be
\int_{arctan(2)}^{arctan(3)}\int_{3csc(\theta)}^{4csc(\theta)}\sqrt{1+4r^2} rdrd\theta.
Finally, for \theta= arctan(3) to arctan(4), the ray enters the rectangle through the left edge, x= r cos(\theta)= 1 (so r= 1/cos(\theta)= sec(\theta)). That integral will be
\int_{arctan(3)}^{arctan(4)}\int_{3sec(\theta)}^{4csc(\theta)}\sqrt{1+4r^2} rdrd\theta.
Now, can you do those integrals? Would you consider that "closed form"?
