If `f'(x) = x-3/x^2,f(1)=11/2`, then f(x)=……..

Answer

To solve the problem, we need to find the function \( f(x) \) given its derivative \( f'(x) = \frac{x - 3}{x^2} \) and the initial condition \( f(1) = \frac{11}{2} \).

### Step-by-Step Solution:

1. **Write down the derivative:**
\[ f'(x) = \frac{x - 3}{x^2} \]

2. **Separate the variables for integration:**
\[ dy = \frac{x - 3}{x^2} dx \]

3. **Rewrite the right-hand side:**
\[ dy = \left( \frac{x}{x^2} - \frac{3}{x^2} \right) dx = \left( \frac{1}{x} - \frac{3}{x^2} \right) dx \]

4. **Integrate both sides:**
\[ \int dy = \int \left( \frac{1}{x} - \frac{3}{x^2} \right) dx \]
The left side integrates to \( y \). The right side integrates as follows:
\[ \int \frac{1}{x} dx = \ln |x| \quad \text{and} \quad \int -\frac{3}{x^2} dx = 3 \cdot \frac{1}{x} \]
Thus, we have:
\[ y = \ln |x| + \frac{3}{x} + C \]

5. **Substitute back for \( f(x) \):**
\[ f(x) = \ln |x| + \frac{3}{x} + C \]

6. **Use the initial condition to find \( C \):**
Given \( f(1) = \frac{11}{2} \):
\[ f(1) = \ln |1| + \frac{3}{1} + C = 0 + 3 + C = \frac{11}{2} \]
Therefore:
\[ 3 + C = \frac{11}{2} \]
\[ C = \frac{11}{2} - 3 = \frac{11}{2} - \frac{6}{2} = \frac{5}{2} \]

7. **Final expression for \( f(x) \):**
Substitute \( C \) back into the equation:
\[ f(x) = \ln |x| + \frac{3}{x} + \frac{5}{2} \]

### Final Answer:
\[ f(x) = \ln |x| + \frac{3}{x} + \frac{5}{2} \]