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 \( I = \int_{2}^{4} \frac{\sqrt{x^2 - 4}}{x^4} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start by rewriting the integrand:
\[ I = \int_{2}^{4} \frac{\sqrt{x^2 - 4}}{x^4} \, dx \]
Notice that 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 integral becomes:
\[ I = \int_{2}^{4} \frac{x \sqrt{1 - \frac{4}{x^2}}}{x^4} \, dx = \int_{2}^{4} \frac{\sqrt{1 - \frac{4}{x^2}}}{x^3} \, dx \]

### Step 2: Substitution
Let \( t = 1 - \frac{4}{x^2} \). Then we differentiate:
\[ dt = \frac{8}{x^3} \, dx \quad \Rightarrow \quad dx = \frac{x^3}{8} \, dt \]
We also need to change the limits of integration. When \( x = 2 \):
\[ t = 1 - \frac{4}{2^2} = 1 - 1 = 0 \]
When \( x = 4 \):
\[ t = 1 - \frac{4}{4^2} = 1 - \frac{4}{16} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, the integral becomes:
\[ I = \int_{0}^{\frac{3}{4}} \sqrt{t} \cdot \frac{1}{8} \, dt \]

### Step 3: Evaluate the integral
Now we can evaluate the integral:
\[ I = \frac{1}{8} \int_{0}^{\frac{3}{4}} t^{\frac{1}{2}} \, dt \]
The integral of \( t^{\frac{1}{2}} \) is:
\[ \int t^{\frac{1}{2}} \, dt = \frac{t^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} t^{\frac{3}{2}} \]
Thus, we have:
\[ I = \frac{1}{8} \cdot \frac{2}{3} \left[ t^{\frac{3}{2}} \right]_{0}^{\frac{3}{4}} = \frac{1}{12} \left[ \left( \frac{3}{4} \right)^{\frac{3}{2}} - 0 \right] \]

### Step 4: Calculate the limits
Now we calculate \( \left( \frac{3}{4} \right)^{\frac{3}{2}} \):
\[ \left( \frac{3}{4} \right)^{\frac{3}{2}} = \frac{3^{\frac{3}{2}}}{4^{\frac{3}{2}}} = \frac{3\sqrt{3}}{8} \]
So substituting back, we get:
\[ I = \frac{1}{12} \cdot \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{96} = \frac{\sqrt{3}}{32} \]

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