Find the following integral: $\int \frac{dx}{\sqrt{5x^{2}-2x}}$

We have $\int \frac{dx}{\sqrt{5x^{2}-2x}} = \int \frac{dx}{\sqrt{5(x^{2}-\frac{2x}{5})}}$.
$= \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{(x-\frac{1}{5})^{2}-(\frac{1}{5})^{2}}}$ (completing the square).
Let $t = x - \frac{1}{5}$,then $dx = dt$.
Therefore,$\int \frac{dx}{\sqrt{5x^{2}-2x}} = \frac{1}{\sqrt{5}} \int \frac{dt}{\sqrt{t^{2}-(\frac{1}{5})^{2}}}$.
Using the standard formula $\int \frac{dx}{\sqrt{x^{2}-a^{2}}} = \log |x + \sqrt{x^{2}-a^{2}}| + C$,we get:
$= \frac{1}{\sqrt{5}} \log |t + \sqrt{t^{2}-(\frac{1}{5})^{2}}| + C$.
Substituting $t = x - \frac{1}{5}$ back,we get:
$= \frac{1}{\sqrt{5}} \log |x - \frac{1}{5} + \sqrt{x^{2}-\frac{2x}{5}}| + C$.