How would you go about to solve it for b)? I tried by parts but its way to long to be the most efficient way to go about.
Calculate the multiple integrals $\displaystyle\iint_{D} f(x,y)\,dy\,dx$ where $f\colon \mathbb{R}^{2}\to\mathbb{R}$ and $D\subset \mathbb{R}^{2}$ is given by:
$a)$ $f(x,y) = x^{3} + 3x^{2}y+y^{3} \text{ and } D = [0,2]\times [0,1]$.
$b)$ $f(x,y) = \displaystyle\frac{x}{x^{2}+y^{2}} \text{ and } D = [1,2]\times [1,2]$.
On
We could use polar coordinates to simplify $f(x,y)$ but also we have to use it in the square.
$$f(r, \theta ) = cos (\theta)/r$$
For the region, we can define
$$r=2sec( \theta), -\pi/4<\theta < \pi/4 $$
$$r=2csc( \theta), \pi/4<\theta <3/ \pi $$
$$r= sec( \theta), 3\pi/4<\theta < 5\pi/4$$
$$r= csc( \theta), 5\pi/4<\theta < 7\pi/4 $$
Step 1: integrate wrt $y$.
$$I=\int_1^2\int_1^2\frac{x}{x^2+y^2}\,dy\,dx=\int_1^2\left[\arctan\left(\frac yx\right)\right]_1^2\,dx=\int_1^2\arctan\left(\frac2x\right)-\arctan\left(\frac1x \right)\,dx$$
Step 2: integrate each term by parts.
$$\int_1^2\arctan\left(\frac2x\right)\,dx=\left[x\arctan\frac2x\right]_1^2+\int_1^2\frac{x}{4+x^2}\,dx\\=2\cdot\frac\pi4-\arctan 2+\log8-{\log{5}}$$
$$\int_1^2\arctan\left(\frac1x\right)\,dx=\left[x\arctan\frac1x\right]_1^2+\int_1^2\frac{x}{1+x^2}\,dx\\=-\frac\pi4+2\arctan \frac12+\frac12\log5-\frac12{\log{2}}$$ So the overall solution is $$\bbox[5px,border:2px solid black]{I=\frac{3\pi}{4}-\arctan 2-2\arctan \left(\frac12\right)+\log\left(\frac85\right)-\frac12\log\left(\frac52\right)}$$