Answer
$$\int \frac{1}{x^2\sqrt{4x-9}}dx=\frac{\sqrt{4x-9}}{9x}+\frac{4}{27}\tan^{-1}\sqrt{\frac{4x-9}{9}}+C$$
Work Step by Step
$$A=\int \frac{1}{x^2\sqrt{4x-9}}dx$$
Using Formula 31, which states that
$$\int \frac{dx}{x^2\sqrt{ax+b}}=-\frac{\sqrt{ax+b}}{bx}-\frac{a}{2b}\int\frac{dx}{x\sqrt{ax+b}}+C$$
with $a=4$ and $b=-9$ here, we have
$$A=-\frac{\sqrt{4x-9}}{-9x}-\frac{4}{-18}\int\frac{dx}{x\sqrt{4x-9}}+C$$ $$A=\frac{\sqrt{4x-9}}{9x}+\frac{2}{9}\int\frac{dx}{x\sqrt{4x-9}}+C$$
For the remaining integral, we apply 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 $a=4$ and $b=9$:
$$A=\frac{\sqrt{4x-9}}{9x}+\frac{2}{9}\Big(\frac{2}{3}\tan^{-1}\sqrt{\frac{4x-9}{9}}\Big)+C$$ $$A=\frac{\sqrt{4x-9}}{9x}+\frac{4}{27}\tan^{-1}\sqrt{\frac{4x-9}{9}}+C$$
