I have the following integral $$\int_0^1 \int_0^1 \frac{xdxdy}{x^2 + y^2} $$
I first took the partial integral with respect to y to obtain the following: $$ \left [ \frac{1}{x} \tan^{-1}x \frac{y}{x} \right ]_{y = 0}^{y = 1} \rightarrow \int_0^1 \frac{1}{x} \tan^{-1}x \, dx \ \text{(now taking partial integral with respect to x)} $$ I try u substitution next but I get stuck.
Also is $$\int_0^1 \int_0^1 \frac{xdydx}{x^2 + y^2} = \int_0^1 \int_0^1 \frac{xdxdy}{x^2 + y^2} ?$$ by Fubuni's Theorem
Your primitive is wrong. The correct one is $$ \int \frac{x}{x^2+y^2}\,dy=\arctan(y/x)+C. $$ This means that you end up calculating $$ \int_0^1\arctan(1/x)\,dx, $$ which is doable.
I first did not see your second question. Yes, you can switch order here. I suggest that you try to do the problem integrating in that order once you are done with the first way. It is a good practice.
Partial integration yields $$\int_{y=0}^{y=1}\frac{x\,dy}{x^2+y^2}=x\int_{y=0}^{y=1}\frac{dy}{x^2+y^2}=x\left[\frac{1}{x}\tan^{-1}\left(\frac{y}{x}\right)\right]_{y=0}^{y=1}=\tan^{-1}\left(\frac{1}{x}\right)$$
$$\int \frac{x}{x^2+y^2}\,dy=x\int\frac{1}{x^2++y^2}\,dy=\frac{1}{x}\int\frac{1}{\frac{y^2}{x^2}+1}\,dy$$
Then let $t:=\frac yx$.
Now that you got your primitive correctly you can procceed.
