Fubini's Theorem solving an integral I've been given the following integral
$\displaystyle \underset{0}{\overset{1}{\int}}\underset{0}{\overset{1}{\int}} \frac{y}{\left(1+x^{2}+y^{2}\right)^{\frac{3}{2}}}dxdy$
And the topic of the exercise is using Fubini's theorem to solve it, I was able to reach a strange integral after swapping the integration order but I have a feeling I'm missing something. Does anyone have an approach I should use instead of brute-forcing it?
This is the outcome of the swap :
$ \displaystyle \underset{0}{\overset{1}{\int}}\frac{1}{\sqrt{1+x^{2}}}-\frac{1}{\sqrt{2+x^{2}}}dx$
EDIT:
So after all this help, I was able to finish the integral properly, my main take on this is to remember the quick substitution which really come in handy in cases like this.
Im adding my final answer if someone ever was wondering :
$\displaystyle \underset{0}{\overset{1}{\int}}\frac{1}{\sqrt{1+x^{2}}}dx-\underset{0}{\overset{1}{\int}}\frac{1}{\sqrt{2+x^{2}}}dx=\ln\left(\sqrt{2}+1\right)-\ln\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)=\ln\left(\frac{\sqrt{2}+1}{\frac{\sqrt{3}+1}{\sqrt{2}}}\right)=\ln\left(\frac{\sqrt{2}+2}{\sqrt{3}+1}\right)$
 A: The integrand function is continuous, then we can use Fubini, then
$$\int_{0}^{1}\int_{0}^{1}\frac{y}{(1+x^2+y^2)^{3/2}}\, dxdy=\int_{0}^{1}\int_{0}^{1}\frac{y}{(1+x^2+y^2)^{3/2}}\, dydx$$
The last double integral is the same that $$\int_{0}^{1}\left(\frac{1}{\sqrt{1+x^2}}-\frac{1}{\sqrt{2+x^2}}\right)\, dx$$
as you said. Then for the integrand $x\mapsto \frac{1}{\sqrt{x^2+a^2}}$ we can use the substitution $u(x)=a\tan(x)$. Eventually then some algebra we arrived to thr value
$$\log(1+\sqrt{2})-\frac{\log(2)-2\log(1+\sqrt{3})}{2}$$
NB: The appearance of the hyperbolic functions that you mention in the comments seems to be the product of some symbolic software, but we don't need those functions. If you need more details, I am happy to write them, just let me know.
A: My hint is just a standard calculus II strategy. I say in my notes (unpublished) that I wrote for my students this past semester, "If you see $\sqrt{x^2 + a^2},$ think $x = a \tan \theta.$"
So we'll do the first of the two integrals and I'll leave the second for you - it follows for a different value of $a$, but the computation is essentially the same.
Let $x = \tan \theta.$ Then $dx = \sec^2 \theta \ d\theta.$ Now
$$\int_0 ^1 \frac{1}{\sqrt{1+x^2}} \ dx = \int_0 ^{\frac{\pi}{4}} \frac{\sec^2 \theta \ d\theta}{\sqrt{1 + \tan^2 \theta}}$$
Now this reduces to the computation of
$$\int_0 ^{\frac{\pi}{4}} \sec \theta \ d\theta$$
using the fact that $\int \sec x \ dx = \ln|\sec x + \tan x| + C.$
