Hello it is my first post here!
I have integral $$ \iint\frac{x^2}{x^2+y^2}\,\mathrm dx\mathrm dy $$ over the area bounded by $y=x$, $2y=x^2$.
I tried to draw the area and it seemed like a pretty straightforward integral. Integrating by $\mathrm dx$, I have that $x\in[0,2]$, $y \in [\frac{x^2}{2}, x]$.
The integral then is:
$$\int_0^2 \int_{\frac{x^2}{2}}^x \frac{x^2}{x^2+y^2} \,\mathrm dy\mathrm dx$$ (swapped the integrals because I can't integrate by dx first since it's in a boundary) which evaluates to $2 - \frac{\pi}{2}$. However my workbook says that the solution is $\ln{2}$. Is there an error?
I verify that your calculation of the iterated integral is correct:
$$ \begin{aligned} &\int_{0}^{2} \int_{\frac{x^{2}}{2}}^{x} \frac{x^{2}}{x^{2}+y^{2}} d y d x =\int_{0}^{2} x\left(\frac{\pi}{4}-\arctan \left(\frac{x}{2}\right)\right) d x =\frac{\pi}{2}-\pi+2=2-\frac\pi2 \end{aligned} $$ where in the second integral, one can use integration by parts.
It is worth knowing that the region that you are integrating on can be also written as $$ D=\{(x,y): 0\le y\le 2\;, y\le x\le \sqrt{2y}\} $$ which gives you the iterated integral: $$ \int_{0}^{2} \int_{y}^{\sqrt{2 y}} \frac{x^{2}}{x^{2}+y^{2}} d x d y $$
So in principle, you can integrate with respect to $x$ first.