Double integral application I need to determine $$\int_{0}^{1} \int_{-\sqrt{x}}^{\sqrt{x}}\frac{1}{1-y}dydx$$
I integrate in terms of the y component and obtained: $$\int_{0}^{1}\ln(\frac{1+\sqrt{x}}{1-\sqrt{x}})dx$$
Can someone help me with the integral in terms of the x component ?
Thank you
 A: Note that $$\ln \left(\dfrac{1+\sqrt{x}}{1-\sqrt{x}}\right) = \ln(1+\sqrt{x}) - \ln(1-\sqrt{x})$$
$$\underbrace{\int \ln(1\pm\sqrt{x})dx = \int \ln(1 \pm t)2tdt}_{x = t^2} = \dfrac12 \left(2(t^2-1)\log(1\pm t)-t(t\mp 2)\right) + \text{constant}$$
Another way would be to integrate over $x$ first.
We have
$$\int_0^1 \int_{-\sqrt{x}}^{\sqrt{x}} \dfrac{dy}{1-y} dx = \int_{y=-1}^1 \int_{x=y^2}^1 \dfrac{dx}{1-y}dy = \int_{y=-1}^1 \dfrac{1-y^2}{1-y} dy = \int_{y=-1}^1 (1+y) dy = 2$$

EDIT
Here is a figure explaining the order of integration and the corresponding limits. The area you are integrating over is the parabola given by $x=y^2$ and the boundary on the right hand side $x=1$.
The way you are integrating you integrate along the yellow strip first and move the yellow strip from $x=0$ to $x=1$. Instead, integrate along the red strip and move the red strip from $y=-1$ to $y=1$.

A: When confronted with a double integral, you should always consider the possibility of switching the order of integration. This problem becomes much easier if you do so. There are many (often, admittedly, artificial) problems that are undoable in one order and easy in the other.
