Answer
$\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
We will use polar coordinates as follows: $x=r \cos \theta; y= r \sin \theta$
We need to work out with the line integral and evaluate the integrand of the double integral as follows:
Consider $I= \iint_{D}(\dfrac{\partial (\tan^{-1} x)}{\partial x}-\dfrac{\partial (\sqrt {x^2+1})}{\partial y})\ dA =\iint_{C} \sqrt {x^2+1} dx+\tan^{-1} x \ dy$
$=- \int_{0}^{1} \int_{x}^{1} (\dfrac{1}{1+x^2}-0) \ dy \ dx $
$= \int_{0}^{1} \dfrac{1}{1+x^2}-\dfrac{x}{1+x^2} \ dx $
$= [\tan^{-1} x-\dfrac{\ln (1+x^2)}{2}]_0^1$
$=\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$
