Question:
Evaluate the following integrals:
$\int \frac{1}{\sqrt{5 x^{2}-2 x}} d x$
Solution:
We have $\int \frac{\mathrm{dx}}{\sqrt{5 \mathrm{x}^{2}-2 \mathrm{x}}}=\int \frac{\mathrm{dx}}{\sqrt{5\left(\mathrm{x}^{2}-\frac{2 \mathrm{x}}{5}\right)}}$
$=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}}$ completing the square
Put $x-1 / 5=t$ then $d x=d t$
Therefore $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{(t)^{2}-\left(\frac{1}{5}\right)^{2}}}$
$=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+c$
$=\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+c$