<s>[INST] How do you find the integral of $\int_a^bdx\int_a^b \frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy$ $$\int_a^bdx\int_a^b \frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy$$
I've tried using the trigonometric substitution $x=\cos t, y=\sin t$ but it didn't work out
 [/INST] By symmetry,
$$ I=\iint_{(a,b)^2}\frac{y^2}{(x^2+y^2)^{3/2}}\,d\mu = \iint_{(a,b)^2}\frac{x^2}{(x^2+y^2)^{3/2}}\,d\mu = \frac{1}{2}\iint_{(a,b)^2}\frac{d\mu}{\sqrt{x^2+y^2}} \tag{1} $$
hence, assuming $a,b>0$:
$$ I = \frac{1}{2}\int_{a}^{b}\log\left(\frac{b+\sqrt{b^2+y^2}}{a+\sqrt{a^2+y^2}}\right)\,dy \tag{2}$$
and:
$$\boxed{ I = (a+b)\log(1+\sqrt{2})+a\log\left(\frac{a}{b+\sqrt{a^2+b^2}}\right)+b\log\left(\frac{b}{a+\sqrt{a^2+b^2}}\right).}\tag{3} $$
</s>