If `int(dx)/(f(x))=log{f(x)}^(2)+c`, then what is f(x) equal to ?

Options

A. `2x+alpha`
B. `x+alpha`
C. `(x)/(2)+alpha`
D. `x^(2)+alpha`

Answer

To solve the equation given in the question, we start with the expression:

\[ \int \frac{dx}{f(x)} = \log(f(x)^2) + c \]

### Step-by-Step Solution:

1. **Differentiate Both Sides**:
To find \( f(x) \), we can differentiate both sides with respect to \( x \).

\[ \frac{d}{dx} \left( \int \frac{dx}{f(x)} \right) = \frac{d}{dx} \left( \log(f(x)^2) + c \right) \]

By the Fundamental Theorem of Calculus, the left-hand side becomes \( \frac{1}{f(x)} \).

The right-hand side can be differentiated using the chain rule:
\[ \frac{d}{dx} \left( \log(f(x)^2) \right) = \frac{2f'(x)}{f(x)} \]

Thus, we have:
\[ \frac{1}{f(x)} = \frac{2f'(x)}{f(x)} \]

2. **Simplify the Equation**:
We can multiply both sides by \( f(x) \) (assuming \( f(x) \neq 0 \)):
\[ 1 = 2f'(x) \]

3. **Solve for \( f'(x) \)**:
Rearranging gives us:
\[ f'(x) = \frac{1}{2} \]

4. **Integrate to Find \( f(x) \)**:
Now, we integrate \( f'(x) \):
\[ f(x) = \frac{1}{2}x + C \]
where \( C \) is a constant of integration.

5. **Conclusion**:
Therefore, the function \( f(x) \) is:
\[ f(x) = \frac{x}{2} + C \]

### Final Answer:
\[ f(x) = \frac{x}{2} + C \]