Integrate $\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y)$ This is question from Axler's Measure, Integration, & Real Analysis Problem 5.B.1, probability measures.

Let $\lambda$ denote Lebesgue measure on $[0,1]$. Show that
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y) = \frac{\pi}{4}$$ and
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(y) d\lambda(x) = -\frac{\pi}{4}.$$

How does one calculate this integral? Since the answer is $\dfrac{\pi}{4}$ is there is a trig substitution  involved?
I also graphed this function and don't understand what the meaning of the graph is from a geometric point of view. 

Note that $$\frac {x^2-y^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y}  \left(\frac{y}{x^2 + y^2}\right) = -\frac{\partial}{\partial x}  \left(\frac{x}{x^2 + y^2}\right),  $$
Riemann and Lebesgue integrals coincide on finite intervals when the integrand is Riemann itegrable.
We can evaluate the inner integral as a Riemann integral using FTC to get for $y \in (0,1]$,
$$F(y) = \int_{[0,1]}\frac {x^2-y^2}{(x^2+y^2)^2}\, d\lambda(x)=  \int_0^1 \frac {x^2-y^2}{(x^2+y^2)^2} \, dx = \left.\frac{-x}{x^2 + y^2}\right|_0^1= \frac{-1}{1+y^2}$$
Now $F$ is integrable over $(0,1]$ and easily evaluated as
$$\int_{[0,1]} F(y) \, d\lambda(y) = \int_0^1\frac{-1}{1+y^2} \, dy = -\arctan(1) = - \frac{\pi}{4}$$
