<s>[INST] Evaluate a double integral. 
To find:

$$I =\int\int_Rx(1+y^2)^{\frac{-1}{2}}dA$$

R is the region in the first quadrant enclosed by $y=x^2$, $y=4$, and $x=0$

$$y=x^2, y=4,x=0, (x= y^\frac{1}{2})$$
$$R=((x,y), 0 \le y \le 4, x^2 \le y \le 4)$$

i.e.

$$R=((x,y), 0 \le x \le 2, x^2 \le y \le 4)$$
$$I=\int_0^2\int_{x^2}^4x(1+y^2)^{\frac{-1}{2}}\,dy\,dx$$

Where to from here?

 [/INST] As you defined well,
$$
R=\left\{ (x,y) \in \mathbb{R}^2: 0\leq x \leq 2, x^2 \leq y \leq 4 \right\}.
$$
So
$$
I =
\int_{R} \frac{x}{\sqrt{1+y^2}} \mathrm{d}A =
\int_0^2 \left( \int_{x^2}^4 \frac{x}{\sqrt{1+y^2}} \mathrm{d}y \right) \mathrm{d}x
$$
Here, either you do a change of variables or you check in any table that:
$$
\frac{\mathrm{d}}{\mathrm{d}z}\sinh^{-1}(z)=\frac{1}{\sqrt{1+y^2}}
$$
So
\begin{align}
I &= \int_0^2 \left[ x\sinh^{-1}(y) \right] ^{y=4}_{y=x^2} \mathrm{d}x \\ 
 &= \int_0^2 \left( x\sinh^{-1}(4) -x\sinh^{-1}(x^2)\right) \mathrm{d}x = \\
&= \left[\frac{x^2}{2}\sinh^{-1}(4)\right]^{x=0}_{x=2} - \int_0^2x\sinh^{-1}(x^2) \mathrm{d}x \\
&= 2\sinh^{-1}(4) - \frac{1}{2}\left[x^2 \sinh^{-1} \left(x^2\right) -\sqrt{x^4+1}\right]^{x=0}_{x=2}\\
&= 2\sinh^{-1}(4) - \frac{1}{2} \left(4 \sinh ^{-1}(4)+1-\sqrt{17}\right) \\
&= \frac{\sqrt{17}-1}{2}.
\end{align}
</s>