The value of the integral `I=int_(1)^(oo) (x^(2)-2)/(x^(3)sqrt(x^(2)-1))dx`, is

Options

A. 0
B. `(2)/(3)`
C. `(4)/(3)`
D. none of these

Answer

To solve the integral \( I = \int_{1}^{\infty} \frac{x^2 - 2}{x^3 \sqrt{x^2 - 1}} \, dx \), we will use the substitution method and properties of definite integrals.

### Step 1: Substitution
Let \( x = \frac{1}{t} \). Then, we differentiate to find \( dx \):
\[ dx = -\frac{1}{t^2} \, dt \]
Next, we change the limits of integration. When \( x = 1 \), \( t = 1 \) and when \( x \to \infty \), \( t \to 0 \).

### Step 2: Change the Integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ I = \int_{1}^{0} \frac{\left(\frac{1}{t}\right)^2 - 2}{\left(\frac{1}{t}\right)^3 \sqrt{\left(\frac{1}{t}\right)^2 - 1}} \left(-\frac{1}{t^2}\right) dt \]
This simplifies to:
\[ I = \int_{0}^{1} \frac{\frac{1}{t^2} - 2}{\frac{1}{t^3} \sqrt{\frac{1}{t^2} - 1}} \frac{1}{t^2} dt \]

### Step 3: Simplifying the Expression
We can simplify the expression inside the integral:
\[ I = \int_{0}^{1} \frac{1 - 2t^2}{\frac{1}{t^4} \sqrt{1 - t^2}} dt \]
This becomes:
\[ I = \int_{0}^{1} \frac{(1 - 2t^2) t^4}{\sqrt{1 - t^2}} dt \]

### Step 4: Factor Out the Negative Sign
Now, we can factor out the negative sign:
\[ I = -\int_{0}^{1} \frac{1 - 2t^2}{\sqrt{1 - t^2}} dt \]

### Step 5: Change the Limits
Changing the limits of integration gives:
\[ I = \int_{0}^{1} \frac{1 - 2t^2}{\sqrt{1 - t^2}} dt \]

### Step 6: Evaluate the Integral
Now, we can split the integral:
\[ I = \int_{0}^{1} \frac{1}{\sqrt{1 - t^2}} dt - 2 \int_{0}^{1} \frac{t^2}{\sqrt{1 - t^2}} dt \]
The first integral evaluates to:
\[ \int_{0}^{1} \frac{1}{\sqrt{1 - t^2}} dt = \frac{\pi}{2} \]
For the second integral, we can use the substitution \( t = \sin \theta \):
\[ \int_{0}^{1} \frac{t^2}{\sqrt{1 - t^2}} dt = \int_{0}^{\frac{\pi}{2}} \sin^2 \theta d\theta = \frac{\pi}{4} \]

### Step 7: Combine Results
Putting it all together:
\[ I = \frac{\pi}{2} - 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{0} \]