If `f'(x)=x/2 + 2/x` and `f(1)=5/4`, then find `f(x)`.

Answer

To find the function \( f(x) \) given that \( f'(x) = \frac{x}{2} + \frac{2}{x} \) and \( f(1) = \frac{5}{4} \), we will follow these steps:

### Step 1: Integrate \( f'(x) \)

We start with the derivative:
\[ f'(x) = \frac{x}{2} + \frac{2}{x} \]

To find \( f(x) \), we need to integrate \( f'(x) \):
\[ f(x) = \int \left( \frac{x}{2} + \frac{2}{x} \right) dx \]

### Step 2: Perform the integration

We can split the integral:
\[ f(x) = \int \frac{x}{2} \, dx + \int \frac{2}{x} \, dx \]

Calculating each integral:
1. For \( \int \frac{x}{2} \, dx \):
\[ \int \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \]

2. For \( \int \frac{2}{x} \, dx \):
\[ \int \frac{2}{x} \, dx = 2 \ln |x| \]

Combining these results, we have:
\[ f(x) = \frac{x^2}{4} + 2 \ln |x| + C \]

### Step 3: Use the initial condition to find \( C \)

We know that \( f(1) = \frac{5}{4} \). So we substitute \( x = 1 \) into our function:
\[ f(1) = \frac{1^2}{4} + 2 \ln |1| + C \]
Since \( \ln |1| = 0 \), this simplifies to:
\[ f(1) = \frac{1}{4} + C \]

Setting this equal to \( \frac{5}{4} \):
\[ \frac{1}{4} + C = \frac{5}{4} \]

### Step 4: Solve for \( C \)

Subtract \( \frac{1}{4} \) from both sides:
\[ C = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1 \]

### Step 5: Write the final function

Substituting \( C \) back into our function:
\[ f(x) = \frac{x^2}{4} + 2 \ln |x| + 1 \]

Thus, the final answer is:
\[ \boxed{f(x) = \frac{x^2}{4} + 2 \ln |x| + 1} \]

---