91Ó°ÊÓ

Given the expression inside the square root, we can recognize the similarity with the Pythagorean identity: $$sin^2(\theta) + cos^2(\theta) = 1$$. This indicates we can use a trigonometric substitution to simplify the square root. Let's choose $$x = 3\sin(\theta)$$, which implies $$dx = 3\cos(\theta)d\theta$$.

Now, let's substitute $$x$$ and $$dx$$ with their expressions in terms of $$\theta$$: $$\int \frac{\sqrt{9-x^{2}}}{x^{2}} d x = \int \frac{\sqrt{9-9\sin^{2}(\theta)}}{(3\sin(\theta))^{2}} \cdot 3\cos(\theta) d\theta$$

Next, we can simplify the integrand. First, we can factor the 9 from the square root and use the Pythagorean identity to further simplify: $$\int \frac{\sqrt{9(1-\sin^{2}(\theta))}}{9\sin^{2}(\theta)} \cdot 3\cos(\theta) d\theta = \int \frac{3\cos(\theta)}{\sin^{2}(\theta)} \cdot 3\cos(\theta) d\theta$$ Multiplying the numerators and denominators, we have: $$\int \frac{9\cos^{2}(\theta)}{\sin^{2}(\theta)} d\theta$$

To proceed, let's use this identity to express the integrand in terms of $$\sin(\theta)$$ only: $$\int \frac{9(1 - \sin^{2}(\theta))}{\sin^{2}(\theta)} d\theta$$

Now, we can separate the integral into two simpler ones: $$\int \frac{9}{\sin^2(\theta)} d\theta - 9\int d\theta$$ The first integral is a well-known integral involving a cosecant function: $$\int \frac{9}{\sin^2(\theta)} d\theta = -9\cot(\theta) + C_1$$ The second integral is just: $$-9\int d\theta =-9\theta + C_2$$ So, the result of the whole integral is: $$-9\cot(\theta) -9\theta + C$$

Finally, we need to express the answer in terms of $$x$$. Recall that we made the substitution $$x = 3\sin(\theta)$$. We can rewrite this as $$\sin(\theta) = \frac{x}{3}$$. To find the angle $$\theta$$, we can use the inverse sine function: $$\theta = \arcsin\left(\frac{x}{3}\right)$$ To find the cotangent function, remember that $$\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$$. Using the Pythagorean identity, we can find $$\cos(\theta)$$: $$\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{x}{3}\right)^2} = \frac{\sqrt{9-x^2}}{3}$$ So, $$\cot(\theta) = \frac{\frac{\sqrt{9-x^2}}{3}}{\frac{x}{3}} = \frac{\sqrt{9-x^2}}{x}$$ Substituting back our expressions for $$\theta$$ and $$\cot(\theta)$$ into our result: $$\int \frac{\sqrt{9-x^{2}}}{x^{2}} d x = -9\frac{\sqrt{9-x^2}}{x} -9\arcsin\left(\frac{x}{3}\right) +C$$