Answer
$\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$
Work Step by Step
Green's Theorem states that:
$\oint_C A\,dx+B \,dy=\iint_{D}(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}) dA $
We need to set up the line integral and compute the integrand of the double integral as follows:
$$ \iint_{D}(\dfrac{\partial (\tan^{-1} x)}{\partial x}-\dfrac{\partial (\sqrt {x^2+1})}{\partial y})dA \\=\iint_{C} \sqrt {x^2+1} dx+\tan^{-1} x \ dy \\=- \int_{0}^{1} \int_{x}^{1} (\dfrac{1}{1+x^2}-0) \ dy \ dx \\= \int_{0}^{1} [\dfrac{1}{1+x^2}-\dfrac{x}{1+x^2}] dx\\=\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$$
