Let f(x) be a function satisfying f'(x) = f(x) and f(0) = 2. Then `int(f(x))/(3+4f(x))dx` is equal to

Options

A. `(1//4) log (3 + 8e^(x)) + C`
B. `(1//2) log (3 + 5e^(x)) + C`
C. `(1//4) log (3 + 4e^(2x)) + C`
D. none of these

Answer

To solve the integral \(\int \frac{f(x)}{3 + 4f(x)} \, dx\) given that \(f'(x) = f(x)\) and \(f(0) = 2\), we can follow these steps:

### Step 1: Find the function \(f(x)\)
Since \(f'(x) = f(x)\), we recognize that this is a standard differential equation. The general solution is:
\[ f(x) = Ce^x \]
where \(C\) is a constant.

### Step 2: Use the initial condition
Given \(f(0) = 2\), we can substitute \(x = 0\) into the equation:
\[ f(0) = Ce^0 = C = 2 \]
Thus, we have:
\[ f(x) = 2e^x \]

### Step 3: Substitute \(f(x)\) into the integral
Now we substitute \(f(x)\) into the integral:
\[ \int \frac{f(x)}{3 + 4f(x)} \, dx = \int \frac{2e^x}{3 + 4(2e^x)} \, dx = \int \frac{2e^x}{3 + 8e^x} \, dx \]

### Step 4: Simplify the integral
We can simplify the integral:
\[ \int \frac{2e^x}{3 + 8e^x} \, dx \]
Let \(t = 8e^x\), then \(dt = 8e^x \, dx\) or \(dx = \frac{dt}{8e^x} = \frac{dt}{t}\) (since \(e^x = \frac{t}{8}\)):
\[ dx = \frac{dt}{t} \]
Now, substituting \(e^x = \frac{t}{8}\) into the integral gives:
\[ \int \frac{2 \cdot \frac{t}{8}}{3 + t} \cdot \frac{dt}{t} = \int \frac{2}{8(3 + t)} \, dt = \frac{1}{12} \int \frac{1}{3 + t} \, dt \]

### Step 5: Evaluate the integral
The integral \(\int \frac{1}{3 + t} \, dt\) is:
\[ \int \frac{1}{3 + t} \, dt = \ln |3 + t| + C \]
Thus, we have:
\[ \frac{1}{12} \ln |3 + t| + C = \frac{1}{12} \ln |3 + 8e^x| + C \]

### Final Answer
The final answer for the integral is:
\[ \int \frac{f(x)}{3 + 4f(x)} \, dx = \frac{1}{12} \ln(3 + 8e^x) + C \]