The value of the integral `int_(2)^(4) (sqrt(x^(2)-4))/(x^(4))dx` is

Options

A. `sqrt((3)/(32))`
B. `(sqrt(3))/(32)`
C. `(32)/(sqrt(3))`
D. `-(sqrt(3))/(32)`

Answer

To solve the integral \( \int_{2}^{4} \frac{\sqrt{x^2 - 4}}{x^4} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We start with the integral:
\[ \int_{2}^{4} \frac{\sqrt{x^2 - 4}}{x^4} \, dx \]
We can factor out \( x^2 \) from the square root:
\[ \sqrt{x^2 - 4} = \sqrt{x^2(1 - \frac{4}{x^2})} = x\sqrt{1 - \frac{4}{x^2}} \]
Thus, the integrand becomes:
\[ \frac{\sqrt{x^2 - 4}}{x^4} = \frac{x\sqrt{1 - \frac{4}{x^2}}}{x^4} = \frac{\sqrt{1 - \frac{4}{x^2}}}{x^3} \]
Now the integral is:
\[ \int_{2}^{4} \frac{\sqrt{1 - \frac{4}{x^2}}}{x^3} \, dx \]

### Step 2: Substitution
Let \( t = 1 - \frac{4}{x^2} \). Then we differentiate \( t \):
\[ dt = \frac{d}{dx}\left(1 - \frac{4}{x^2}\right) = 0 + \frac{8}{x^3} \, dx = \frac{8}{x^3} \, dx \]
Thus, we have:
\[ dx = \frac{x^3}{8} dt \]

### Step 3: Change the limits of integration
When \( x = 2 \):
\[ t = 1 - \frac{4}{2^2} = 1 - 1 = 0 \]
When \( x = 4 \):
\[ t = 1 - \frac{4}{4^2} = 1 - \frac{1}{4} = \frac{3}{4} \]
So the limits change from \( x: 2 \to 4 \) to \( t: 0 \to \frac{3}{4} \).

### Step 4: Substitute into the integral
Now substituting into the integral:
\[ \int_{0}^{\frac{3}{4}} \sqrt{t} \cdot \frac{x^3}{8} \cdot \frac{1}{x^3} dt = \int_{0}^{\frac{3}{4}} \frac{\sqrt{t}}{8} dt \]
This simplifies to:
\[ \frac{1}{8} \int_{0}^{\frac{3}{4}} \sqrt{t} \, dt \]

### Step 5: Evaluate the integral
The integral \( \int \sqrt{t} \, dt \) is:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \]
Now we evaluate it from \( 0 \) to \( \frac{3}{4} \):
\[ \frac{1}{8} \left[ \frac{2}{3} t^{3/2} \right]_{0}^{\frac{3}{4}} = \frac{1}{8} \cdot \frac{2}{3} \left( \left( \frac{3}{4} \right)^{3/2} - 0 \right) \]
Calculating \( \left( \frac{3}{4} \right)^{3/2} \):
\[ \left( \frac{3}{4} \right)^{3/2} = \frac{3^{3/2}}{4^{3/2}} = \frac{3\sqrt{3}}{8} \]
So we have:
\[ \frac{1}{8} \cdot \frac{2}{3} \cdot \frac{3\sqrt{3}}{8} = \frac{1}{8} \cdot \frac{2 \cdot 3\sqrt{3}}{24} = \frac{3\sqrt{3}}{192} = \frac{\sqrt{3}}{64} \]

### Final Answer
Thus, the value of the integral is:
\[ \frac{\sqrt{3}}{32} \]