Find the indefinite integral (Remember to use adsolute values wnere appropriate. Us $$ \int \frac{\sqrt{64-x^{2}}}{x} d x $$ $$ \square $$

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We start with the integral
$$
I=\int \frac{\sqrt{64-x^{2}}}{x}\,dx.
$$
A standard approach for integrals containing $$\sqrt{a^2-x^2}$$ is to use the substitution
$$
x=8\sin\theta, \quad \text{so that} \quad dx=8\cos\theta\,d\theta.
$$
Then, we have
$$
\sqrt{64-x^2}=\sqrt{64-64\sin^2\theta}=\sqrt{64\cos^2\theta}=8\lvert\cos\theta\rvert.
$$
For $$\theta$$ chosen in the interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$, where $$\cos\theta\ge0$$, we have $$\lvert\cos\theta\rvert=\cos\theta$$.

Substituting into the integral, we get
$$
I=\int \frac{8\cos\theta}{8\sin\theta}\, (8\cos\theta\,d\theta)
=\int \frac{8\cos\theta\cdot 8\cos\theta}{8\sin\theta}\,d\theta
=\int \frac{64\cos^2\theta}{8\sin\theta}\,d\theta
=8\int \frac{\cos^2\theta}{\sin\theta}\,d\theta.
$$

Express $$\cos^2\theta$$ in terms of $$\sin\theta$$ using the identity
$$
\cos^2\theta=1-\sin^2\theta:
$$
$$
8\int \frac{1-\sin^2\theta}{\sin\theta}\,d\theta
=8\left[\int \frac{1}{\sin\theta}\,d\theta-\int \sin\theta\,d\theta\right].
$$

Now, evaluate each integral separately.

1. The integral of $$\csc\theta$$:
   $$
   \int \csc\theta\,d\theta=\ln\left|\tan\frac{\theta}{2}\right|+C_1.
   $$
   
2. The integral of $$\sin\theta$$:
   $$
   \int \sin\theta\,d\theta=-\cos\theta+C_2.
   $$

Thus, we have
$$
I=8\left[\ln\left|\tan\frac{\theta}{2}\right|+ \cos\theta\right]+C.
$$

We now revert the expression back in terms of $$x$$. Recall our substitution
$$
x=8\sin\theta \quad \Longrightarrow \quad \sin\theta=\frac{x}{8}.
$$
Also, 
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{x^2}{64}}=\frac{\sqrt{64-x^2}}{8}.
$$
To express $$\tan\frac{\theta}{2}$$ in terms of $$x$$, we use the half-angle formula:
$$
\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}.
$$
Substitute the expressions for $$\sin\theta$$ and $$\cos\theta$$:
$$
\tan\frac{\theta}{2}=\frac{1-\frac{\sqrt{64-x^2}}{8}}{\frac{x}{8}}
=\frac{8-\sqrt{64-x^2}}{x}.
$$

Finally, substituting back, the integral becomes
$$
I=8\ln\left|\frac{8-\sqrt{64-x^2}}{x}\right|+ \sqrt{64-x^2}+C.
$$

Thus, the answer is
$$
\boxed{8\ln\left|\frac{8-\sqrt{64-x^2}}{x}\right|+\sqrt{64-x^2}+C}.
$$
The indefinite integral is: $$ 8\ln\left|\frac{8-\sqrt{64-x^2}}{x}\right| + \sqrt{64-x^2} + C $$

Find the indefinite integral (Remember to use adsolute values wnere appropriate. Us

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