`int(dx)/(x^(2)sqrt(16-x^(2)))` has the value equal to

Options

A. `C-(1)/(4)tan^(-1)sec((x)/(4))`
B. `(1)/(4)tan^(-1)sec((x)/(4))+C`
C. `C-(sqrt(16-x^(2)))/(16x)`
D. `(sqrt(16-x^(2)))/(16x)+C`

Explanation

To solve the integral

\[ I = \int \frac{dx}{x^2 \sqrt{16 - x^2}}, \]

we can use a substitution method. Let's go through the steps:

### Step 1: Substitution
Let \( x = 4 \sin \theta \). Then, \( dx = 4 \cos \theta \, d\theta \).

### Step 2: Change of Limits and Simplifying the Integral
Substituting \( x \) into the integral, we have:

\[ \sqrt{16 - x^2} = \sqrt{16 - (4 \sin \theta)^2} = \sqrt{16(1 - \sin^2 \theta)} = \sqrt{16 \cos^2 \theta} = 4 \cos \theta. \]

Now substitute \( x \) and \( dx \) into the integral:

\[ I = \int \frac{4 \cos \theta \, d\theta}{(4 \sin \theta)^2 \cdot 4 \cos \theta} = \int \frac{4 \cos \theta \, d\theta}{16 \sin^2 \theta \cdot 4 \cos \theta}. \]

This simplifies to:

\[ I = \int \frac{d\theta}{16 \sin^2 \theta} = \frac{1}{16} \int \csc^2 \theta \, d\theta. \]

### Step 3: Integrate
The integral of \( \csc^2 \theta \) is:

\[ \int \csc^2 \theta \, d\theta = -\cot \theta + C. \]

Thus, we have:

\[ I = -\frac{1}{16} \cot \theta + C. \]

### Step 4: Back Substitution
Now we need to revert back to \( x \). Recall that:

\[ \sin \theta = \frac{x}{4} \implies \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{16 - x^2}}{x}. \]

So we substitute back:

\[ I = -\frac{1}{16} \cdot \frac{\sqrt{16 - x^2}}{x} + C. \]

### Final Result
Thus, the value of the integral is:

\[ I = -\frac{\sqrt{16 - x^2}}{16x} + C. \]