How to solve this double integral? I can put this equation
$$ \int_0^{\sqrt{2}}\int_{x^2}^2 x (y^2+2)^{\frac{1}{4}} \, dy ~dx$$
into Wolfram but I don't see the intermediate steps and don't know how to solve this directly. I tried u-substitution but that doesn't seem to apply. It might be a hypergeometric but I don't know how to tackle it. 
PS: This is not homework
 A: Perhaps you meant:
$$\int\limits_0^{\sqrt 2}\int\limits_{x^2}^2x\sqrt[4]{y^2+2}dydx=\int\limits_0^2\int\limits_0^{\sqrt y}x\sqrt[4]{y^2+2}dxdy=\frac{1}{2}\int\limits_0^2y\sqrt[4]{y^2+2}\;dy=$$
$$=\frac{1}{4}\int\limits_0^2(2y\,dy)(y^2+2)^{1/4}=\left.\frac{1}{4}\frac{4}{5}(y^2+2)^{5/4}\right|_0^2=\frac{1}{5}\left[6^{5/4}-2^{5/4}\right]$$
A: Draw the curve $y=x^2$, and the line $y=2$, and the line $x=\sqrt{2}$.
Our integral has $y$ going from $x^2$ to $2$, and then $x$ going from $0$ to $\sqrt{2}$.
So we are integrating over the part of the first quadrant which is above $y=x^2$, and below $y=2$.
We don't really want to try to find an antiderivative of $(y^2+1)^{1/4}$. In fact, this function does not have an elementary antiderivative. But at least it looks hard.
So integrate first with respect to $x$, from $x=0$ to $x=\sqrt{y}$. Then let $y$ go from $0$ to $2$.
When we integrate $x$, we get $x^2/2$. When we evaluate at $\sqrt{y}$ and $0$, we end up with an expression of shape $ky(y^2+2)^{1/4}$. This integrates easily.
