Answer
$$\int\frac{dx}{x\sqrt{x-3}}=\frac{2}{\sqrt3}\tan^{-1}\sqrt{\frac{x-3}{3}}+C$$
Work Step by Step
$$A=\int\frac{dx}{x\sqrt{x-3}}$$
Using Formula 29b, which states that
$$\int\frac{dx}{x\sqrt{ax-b}}=\frac{2}{\sqrt b}\tan^{-1}\sqrt{\frac{ax-b}{b}}+C$$
with $a=1$ and $b=3$ here, we have
$$A=\frac{2}{\sqrt3}\tan^{-1}\sqrt{\frac{x-3}{3}}+C$$
