Help to resolve a Double Integral I'm doing a workout guide about double integrals and I came across an exercise that I could not resolve for a while.

$$\int_0^2\int_1^2 \frac{x}{\sqrt{1+x^2+y^2}} \,\mathrm dx\,\mathrm dy$$

I guess that the easier order of integration is $dxdy$, because if I try to integrate respect to $y$ first, I would have to deal with the integral of a root of $1+x^2+y^2$, while outside of the root there's no $y$.

I tried Integration by substitution (with $u = 1 + x^2 + y^2$ and $du = 2xdx$) to solve the inner integral, but then I felt I couldn't solve the outside integral.

Any hints you can give me?.
The antiderivative may be deduced, but it takes a little patience.  I prefer to integrate by parts rather than do a trig sub right away.

$$\begin{align}\int dx \sqrt{x^2+a^2} &amp;= x \sqrt{x^2+a^2} - \int dx \frac{x^2}{\sqrt{x^2+a^2}}\\ &amp;= x \sqrt{x^2+a^2} - \int dx \sqrt{x^2+a^2} + a^2 \int \frac{dx}{\sqrt{x^2+a^2}} \end{align}$$

Now a trig sub is less work:

$$\implies 2 \int dx \sqrt{x^2+a^2} = x \sqrt{x^2+a^2}  + a^2 \int dt \, \sec{t} $$

or


  $$\int dx \sqrt{x^2+a^2} = \frac12 x \sqrt{x^2+a^2}  + \frac12 a^2 \log{\left (x+\sqrt{a^2+x^2}\right )} + C$$
How hard can it be?? I mean once you substitute $x=r\cos\theta$ and $y=r\sin\theta$, all you need to do it find the limits of $r$ and $\theta$.Now $1 \le r\cos\theta \le 2$ and $0 \le r\sin \theta \le 2$. Squaring both sides it is easy to see that $1 \le r \le 2\sqrt{2}$ . Similarly you can find out that $ 0 \le \theta \le tan^{-1}(2)=\theta_{1}$. Then Substituting them in the integral yields
$$\int_{0}^{\theta_1} \int_{1}^{2\sqrt{2}}\frac{r\cos\theta}{\sqrt{1+r^2}} rdrd\theta$$

This can be integrated easily.