University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{\sqrt{4-x^2}}{x}dx=\sqrt{4-x^2}-2\ln\Big|\frac{2+\sqrt{4-x^2}}{x}\Big|+C$$
$$A=\int\frac{\sqrt{4-x^2}}{x}dx$$ Use Formula 47, which states that $$\int \frac{\sqrt{a^2-x^2}}{x}dx=\sqrt{a^2-x^2}-a\ln\Big|\frac{a+\sqrt{a^2-x^2}}{x}\Big|+C$$ for $a=2$ here: $$A=\sqrt{4-x^2}-2\ln\Big|\frac{2+\sqrt{4-x^2}}{x}\Big|+C$$