Evaluating a given contour integral using Green's Theorem 
Let $D$ be the region in $\mathbb{R}^2$ that contains the points $(x,y) : x^2 + y^2 \leq 1$ and $y \geq 0 $. Let $C$ be the curve enclosing $D$ oriented against the clock. Evaluate $$ \int\limits_C \left(xy + \ln(x^2 + 1) \right) dx + \left(4x + e^{y^2} + 3\arctan(y)\right) dy .$$

This is what I've got: 
$$D = \left \{(x,y):-1 \leq x \leq 1, 0 \leq y \leq \sqrt{1 - x^2} \right \}$$
Now if i'm using the fact that: $$\int \limits_C P\:dx + Q\:dy = \iint \limits_D \left(Q_x - P_y \right)\:dxdy = \iint \limits_D \left(4 - x\right)\:dxdy $$
Where $Q_x,\:P_y$ are the partial derivatives. 
Now trying to solve this the way i'vs set this up with $D$ i get some expressions that certainly doesn't look like the answer my book gives me $(2\pi)$. Have i set this one up right and calculated wrong? Or is there a much easier way to set this one up? I can use any method as long as it touches on Green's Theorem, i.e., path integral or double, changing of bounds, etc.
Any help would be appreciated.
Thanks in advance!

Your expression leads to an answer of $2\pi$. An easy way is to convert to the polar form. Then, $x=r\cos\theta$ and $y=r\sin\theta$. The determinant of the Jacobian matrix is $r$, thus, $dxdy=rdrd\theta$.
$$\iint \limits_D \left(4 - x\right)\:dxdy = \int_{\theta=0}^{\pi}\int_{r=0}^{1}(4r-r^2\cos\theta) drd\theta=\int_{\theta=0}^{\pi}\left[2r^2-\frac13r^3\cos\theta\right]_0^1d\theta$$$$=\int_{\theta=0}^{\pi}2-\frac13\cos\theta d\theta=\left[2\theta - \frac13\sin\theta\right]_0^{\pi}=2\pi$$
