Answer
$\dfrac{\sqrt{4x-9}}{9x}+(\dfrac{4}{27}) \tan^{-1} [\sqrt{\dfrac{4x-9}{9}}] +C$
Work Step by Step
Apply the formula:
$\int \dfrac{dx}{x^2\sqrt{ax+b}}=-\dfrac{\sqrt{ax+b}}{bx}-\dfrac{a}{2b}\int\frac{dx}{x\sqrt{ax+b}}+C$ and $\int\dfrac{dx}{x\sqrt{ax-b}}=\dfrac{2}{\sqrt b}\tan^{-1}\sqrt{\dfrac{ax-b}{b}}+C$
Let $I=-\dfrac{\sqrt{4x-9}}{-9x}- (\dfrac{4}{-18}) \times \int\dfrac{dx}{x\sqrt{4x-9}}+C\\=\dfrac{\sqrt{4x-9}}{9x}+\dfrac{2}{9}\int\dfrac{dx}{x\sqrt{4x-9}}+C \\=\dfrac{\sqrt{4x-9}}{9x}+(\dfrac{4}{27}) \tan^{-1} [\sqrt{\dfrac{4x-9}{9}}] +C$
