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?.
 A: 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.
$$\begin{align}\int dx \sqrt{x^2+a^2} &= x \sqrt{x^2+a^2} - \int dx \frac{x^2}{\sqrt{x^2+a^2}}\\ &= x \sqrt{x^2+a^2} - \int dx \sqrt{x^2+a^2} + a^2 \int \frac{dx}{\sqrt{x^2+a^2}} \end{align}$$
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

$$\int dx \sqrt{x^2+a^2} = \frac12 x \sqrt{x^2+a^2}  + \frac12 a^2 \log{\left (x+\sqrt{a^2+x^2}\right )} + C$$

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2}\int_{1}^{2}
{x \over \root{1 + x^{2} + y^{2}}}\,\dd x\,\dd y} =
\int_{0}^{2}\int_{1}^{2}
\nabla\times\pars{\root{1 + x^{2} + y^{2}}\,\hat{y}}
\cdot\dd\vec{S}
\\[5mm] = &\
\oint_{\partial\bracks{\pars{1,2}\times\pars{0,2}}}\,\,\,
\root{1 + x^{2} + y^{2}}\,\hat{y}\cdot\dd\vec{r}
\\[5mm] = &\
\int_{0}^{2}\root{1 + 2^{2} + y^{2}}\,\dd y +
\int_{2}^{0}\root{1 + 1^{2} + y^{2}}\,\dd y
\\[5mm] = &\
\underbrace{\int_{0}^{2}\root{5 + y^{2}}\,\dd y}
_{\ds{3 + {5 \over 4}\ln\pars{5}}}\ -\
\underbrace{\int_{0}^{2}\root{2 + y^{2}}\,\dd y}
_{\ds{\root{6} + \ln\pars{\root{3} + \root{2}}}}
\\[5mm] = &\
\bbx{3 - \root{6} + {5 \over 4}\,\ln\pars{5} -
\ln\pars{\root{3} + \root{2}}} \approx 1.4161 \\ &
\end{align}
A: 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.
