`int(5x-1)/(3x^2+x+2)dx=`

Options

A. `(5)/(6)log|3x^2+x+2|-(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
B. `(5)/(6)log|3x^2+x+2|+(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
C. `(5)/(3)log|3x^2+x+2|-(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`
D. `(5)/(3)log|3x^2+x+2|+(11)/(3sqrt(23))tan^(-1)((6x+1)/(sqrt(23)))+c`

Answer

To solve the integral \(\int \frac{5x - 1}{3x^2 + x + 2} \, dx\), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{5x - 1}{3x^2 + x + 2} = \frac{6x + 1 - (x + 2)}{3x^2 + x + 2} \]
This allows us to separate the fraction:
\[ = \frac{6x + 1}{3x^2 + x + 2} - \frac{x + 2}{3x^2 + x + 2} \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ \int \frac{5x - 1}{3x^2 + x + 2} \, dx = \int \frac{6x + 1}{3x^2 + x + 2} \, dx - \int \frac{x + 2}{3x^2 + x + 2} \, dx \]

### Step 3: Solve the first integral
For the first integral, we can use substitution. Let:
\[ t = 3x^2 + x + 2 \implies dt = (6x + 1) \, dx \]
Thus, the first integral becomes:
\[ \int \frac{6x + 1}{3x^2 + x + 2} \, dx = \int \frac{dt}{t} = \ln |t| + C_1 = \ln |3x^2 + x + 2| + C_1 \]

### Step 4: Solve the second integral
For the second integral, we can simplify it:
\[ \int \frac{x + 2}{3x^2 + x + 2} \, dx \]
This can be approached by partial fraction decomposition or direct integration. We can perform polynomial long division if necessary, but in this case, we can directly integrate:
\[ \int \frac{x + 2}{3x^2 + x + 2} \, dx \]
This integral can be solved by recognizing that the derivative of the denominator is \(6x + 1\), which is related to the numerator.

### Step 5: Combine results
Combining the results from both integrals, we have:
\[ \int \frac{5x - 1}{3x^2 + x + 2} \, dx = \frac{5}{6} \ln |3x^2 + x + 2| - \frac{1}{6} \int \frac{dt}{t} + C \]
The second integral simplifies to:
\[ -\frac{1}{6} \ln |3x^2 + x + 2| + C_2 \]
Thus, we can combine these results:
\[ \int \frac{5x - 1}{3x^2 + x + 2} \, dx = \frac{5}{6} \ln |3x^2 + x + 2| - \frac{1}{6} \ln |3x^2 + x + 2| + C \]
This simplifies to:
\[ \frac{4}{6} \ln |3x^2 + x + 2| + C \]

### Final Answer
The final answer is:
\[ \int \frac{5x - 1}{3x^2 + x + 2} \, dx = \frac{2}{3} \ln |3x^2 + x + 2| + C \]