If `I=int(dx)/(x^(2)sqrt(1+x^(2)))=(f(x))/(x)+C, (AA x gt0)` (where, C is the constant of integration) and `f(1)=-sqrt2`, then the value of `|f(sqrt3)|` is

Answer

To solve the given problem, we need to evaluate the integral \( I = \int \frac{dx}{x^2 \sqrt{1+x^2}} \) and relate it to the function \( f(x) \) given by the equation \( I = \frac{f(x)}{x} + C \), where \( C \) is a constant of integration. We also know that \( f(1) = -\sqrt{2} \), and we need to find \( |f(\sqrt{3})| \).

### Step-by-Step Solution:

1. **Evaluate the Integral**:
We start with the integral:
\[ I = \int \frac{dx}{x^2 \sqrt{1+x^2}} \]
To simplify this integral, we can factor out \( x^2 \) from the square root in the denominator:
\[ I = \int \frac{dx}{x^2 \sqrt{x^2(1+\frac{1}{x^2})}} = \int \frac{dx}{x^2 \cdot x \sqrt{1+\frac{1}{x^2}}} = \int \frac{dx}{x^3 \sqrt{1+\frac{1}{x^2}}} \]

2. **Substitution**:
Let \( t = 1 + \frac{1}{x^2} \). Then, differentiating both sides gives:
\[ dt = -\frac{2}{x^3} dx \quad \Rightarrow \quad dx = -\frac{x^3}{2} dt \]
Substituting \( dx \) into the integral:
\[ I = \int \frac{-\frac{x^3}{2} dt}{x^3 \sqrt{t}} = -\frac{1}{2} \int \frac{dt}{\sqrt{t}} = -\frac{1}{2} \cdot 2\sqrt{t} + C = -\sqrt{t} + C \]

3. **Back Substitution**:
Now, substituting back for \( t \):
\[ I = -\sqrt{1+\frac{1}{x^2}} + C = -\sqrt{\frac{x^2 + 1}{x^2}} + C = -\frac{\sqrt{x^2 + 1}}{x} + C \]

4. **Relate to \( f(x) \)**:
From the problem, we have:
\[ I = \frac{f(x)}{x} + C \]
Therefore, we can equate:
\[ -\frac{\sqrt{x^2 + 1}}{x} + C = \frac{f(x)}{x} + C \]
This implies:
\[ f(x) = -\sqrt{x^2 + 1} \]

5. **Evaluate \( f(1) \)**:
We know \( f(1) = -\sqrt{1^2 + 1} = -\sqrt{2} \), which is consistent with the given condition.

6. **Find \( f(\sqrt{3}) \)**:
Now we calculate \( f(\sqrt{3}) \):
\[ f(\sqrt{3}) = -\sqrt{(\sqrt{3})^2 + 1} = -\sqrt{3 + 1} = -\sqrt{4} = -2 \]

7. **Calculate the Modulus**:
Finally, we need to find \( |f(\sqrt{3})| \):
\[ |f(\sqrt{3})| = |-2| = 2 \]

### Final Answer:
The value of \( |f(\sqrt{3})| \) is \( 2 \).