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

$\frac{1}{\sqrt{5}} \ln \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2}{5}x} \right| + C$

The correct answer is Option (2) → $\frac{1}{\sqrt{5}} \ln \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2}{5}x} \right| + C$

We have $\int \frac{dx}{\sqrt{5x^2 - 2x}} = \int \frac{dx}{\sqrt{5\left(x^2 - \frac{2x}{5}\right)}}$

$= \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2}}$ (completing the square)

Put $x - \frac{1}{5} = t$. Then $dx = dt$.

Therefore, $\int \frac{dx}{\sqrt{5x^2 - 2x}} = \frac{1}{\sqrt{5}} \int \frac{dt}{\sqrt{t^2 - \left(\frac{1}{5}\right)^2}}$

$= \frac{1}{\sqrt{5}} \log \left| t + \sqrt{t^2 - \left(\frac{1}{5}\right)^2} \right| + C$

$= \frac{1}{\sqrt{5}} \log \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2x}{5}} \right| + C$