## University Calculus: Early Transcendentals (3rd Edition)

$$\int \frac{\sqrt{9-4x}}{x^2}dx=-\frac{\sqrt{9-4x}}{x}-\frac{2}{3}\ln\Big|\frac{\sqrt{9-4x}-3}{\sqrt{9-4x}+3}\Big|+C$$
$$A=\int \frac{\sqrt{9-4x}}{x^2}dx$$ Using Formula 30, which states that $$\int \frac{\sqrt{ax+b}}{x^2}dx=-\frac{\sqrt{ax+b}}{x}+\frac{a}{2}\int\frac{dx}{x\sqrt{ax+b}}+C$$ with $a=-4$ and $b=9$ here, we have $$A=-\frac{\sqrt{9-4x}}{x}+\frac{(-4)}{2}\int\frac{dx}{x\sqrt{9-4x}}+C$$ $$A=-\frac{\sqrt{9-4x}}{x}-2\int\frac{dx}{x\sqrt{9-4x}}+C$$ For the remaining integral, we apply Formula 29a, which states that $$\int\frac{dx}{x\sqrt{ax+b}}=\frac{1}{\sqrt b}\ln\Big|\frac{\sqrt{ax+b}-\sqrt b}{\sqrt{ax+b}+\sqrt b}\Big|+C$$ with $a=-4$ and $b=9$: $$A=-\frac{\sqrt{9-4x}}{x}-2\Big(\frac{1}{3}\ln\Big|\frac{\sqrt{9-4x}-3}{\sqrt{9-4x}+3}\Big|\Big)+C$$ $$A=-\frac{\sqrt{9-4x}}{x}-\frac{2}{3}\ln\Big|\frac{\sqrt{9-4x}-3}{\sqrt{9-4x}+3}\Big|+C$$