`int(3x+1/x+4)dx=`

Options

A. `3/2 x^2 +log x+4x+c`
B. `3/2 x^3 +2 log x+4x +c`
C. `3/2 x^3+log x -4x +c`
D. `3/2 x^2 - log x+4x +c`

Answer

To solve the integral \( \int \left( \frac{3x + 1}{x + 4} \right) dx \), we will follow these steps:

### Step 1: Simplify the integrand
We can separate the integrand into simpler parts. We can rewrite the expression as:
\[ \frac{3x + 1}{x + 4} = \frac{3x + 12 - 11}{x + 4} = \frac{3(x + 4) - 11}{x + 4} = 3 - \frac{11}{x + 4} \]
Thus, we can rewrite the integral:
\[ \int \left( \frac{3x + 1}{x + 4} \right) dx = \int \left( 3 - \frac{11}{x + 4} \right) dx \]

### Step 2: Integrate each term
Now we will integrate each term separately:
1. The integral of \( 3 \) is:
\[ \int 3 \, dx = 3x \]
2. The integral of \( -\frac{11}{x + 4} \) is:
\[ -11 \int \frac{1}{x + 4} \, dx = -11 \ln |x + 4| \]

### Step 3: Combine the results
Now we combine the results of the integrals:
\[ \int \left( 3 - \frac{11}{x + 4} \right) dx = 3x - 11 \ln |x + 4| + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:
\[ \int \left( \frac{3x + 1}{x + 4} \right) dx = 3x - 11 \ln |x + 4| + C \]

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