Show the function defined on $[0,1] \times [0,1]$ via $\frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y) \neq 0$ and $0$ otherwise is not integrable. Let $f: [0,1] \times [0,1] \rightarrow \Bbb{R}$ be defined via
$$f(x,y)=
\begin{cases}
\frac{x^2-y^2}{(x^2+y^2)^2} &(x,y) \neq (0,0) \\
(0,0) & \text{otherwise}
\end{cases}$$
So I know if I integrate with respect to $x$ I get
$$\frac{-x}{x^2+y^2}$$
and if I integrate this with respect to $y$ I get
$$-\tan^{-1}(\frac{y}{x})$$
which is bounded. How is this function non-integrable? also when integrating w.r.t. $x$ was I supposed to plug in end points $0$ and $1$ before moving onto integrating $y$?

I claim that
$$\int_0^1 \int_0^1 f(x,y) dx dy \neq \int_0^1 \int_0^1 f(x,y) dy dx.$$
We compute the LHS:
\begin{align*}
    \int_0^1 \int_0^1 f(x,y) dx dy &= \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx dy\\
    &= \int_0^1 -\frac{x}{x^2+y^2} \bigg\vert_0^1 dy\\
    &=-\int_0^1 \frac{1}{1+y^2}dy\\
    &=- \tan^{-1}(y) \bigg \vert_0^1\\
    &= -\frac{\pi}{4}
\end{align*}
Now we compute the RHS:
\begin{align*}
    \int_0^t \int_0^1 f(x,y) dydx &= \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy dx\\
    &=\int_0^1 \frac{y}{x^2+y^2} \bigg \vert_0^1 dx\\
    &= \int_0^1 \frac{1}{x^2+1} dx\\
    &= \tan^{-1}(x) \bigg \vert_0^1 \\
    &= \frac{\pi}{4}
\end{align*}
thus $f(x,y)$ is not integrable.
