Evaluate $\int_{-1}^1 \frac {1}{\sqrt{|x|}} \text{d}x$ I need some help to solve this integral with absolute value. I'm not sure how to do these types of integrals.

$$\int_{-1}^1 \frac {1}{\sqrt{|x|}} \text{d}x$$

Thank you 
 A: Mike Earnest's hint about the symmetrical property of this particular question works, but I will like to present another method/hint on dealing with modulus functions that still works even when dealing with asymmetrical limits.
It involves splitting up the modulus into cases by restricting the domain.
$$ |f(x))| =
\begin{cases}
f(x) & \textrm{if } f(x) \geq 0 \\
-f(x) & \textrm{if } f(x) < 0 
\end{cases}$$
In particular,
$$ |x| =
\begin{cases}
x & \textrm{if } x \geq 0 \\
-x & \textrm{if } x < 0 
\end{cases}$$
Hence, our required integral can be decomposed into
\begin{equation} \int_{-1}^{1} \frac{1}{\sqrt{|x|}} \mathrm{d}x = \int_{-1}^0 \frac{1}{\sqrt{-x}} \mathrm{d}x + \int_{0}^{1} \frac{1}{\sqrt{x}} \mathrm{d}x \end{equation}
As AlexR points out, even though the function isn't defined for $x=0$, the integrals still work by considering limits for some small $\epsilon$. In particular, both integrals on the right exist, so the integral for the question exists and the equality is valid.
