I'm having trouble integrating this arc length.

General formula:

$\displaystyle L=\int{\sqrt{1+(f'(x))^2}} dx$, between intervalsaandb

My function: $\displaystyle y=ax^2-57$, where a=0.00183

So $\displaystyle \frac{dy}{dx}=2ax$

Putting this derivative into general formula:

$\displaystyle L=\int{\sqrt{1+4a^2x^2}} dx$, between the intervals of176.3and-176.3. (I haven't put a=0.00183 in yet because it would get too messy intergrating it.)

I don't know how to integrate this because I think it involves some method of substitution which I don't know. Any help would be great ;D