Answer
$\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$
Work Step by Step
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA$
We set up the line integral and find out the integrand of the double integral as follows:
$\oint_CP\,dx+Q\,dy= \iint_{D}(\dfrac{\partial (\tan^{-1} x)}{\partial x}-\dfrac{\partial (\sqrt {x^2+1})}{\partial y})dA$
or, $=\iint_{C} \sqrt {x^2+1} dx+\tan^{-1} x \ dy$
or, $=- \int_{0}^{1} \int_{x}^{1} (\dfrac{1}{1+x^2}-0) \ dy \ dx $
or, $= \int_{0}^{1} \dfrac{1}{1+x^2}-\dfrac{x}{1+x^2} \ dx $
or, $= [\tan^{-1} x-\dfrac{\ln (1+x^2)}{2}]_0^1$
or, $=\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$
