Question

Find $\int \frac{1}{x\sqrt{x^2-x}}dx$

Answer 1

$\int \frac{1}{x\sqrt{x^2-x}}dx$

$=\int \frac{dx}{x\sqrt{x^2(1-\frac{1}{x})}}$

$=\int \frac{dx}{x^2\sqrt{1-\frac{1}{x}}}$

Now,

Let $1-\frac{1}{x}=t\Rightarrow \frac{dx}{x^2}=dt$

Therefore,

$=\int \frac{dt}{\sqrt{t}}$

$=2\sqrt{t}+C$

$=2\sqrt{1-\frac{1}{x}}+C(\because t=1-\frac{1}{x})$