`int(x+1)/(x^2+5x+6)dx=`

Options

A. `log|((x+3)^2)/(x+2)|+c`
B. `log|(x+2)/((x+3)^2)|+c`
C. `2log|(x+3)/(x+2)|+c`
D. `2log|(x+2)/(x+3)|+c`

Answer

To solve the integral \(\int \frac{x+1}{x^2+5x+6} \, dx\), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Factor the Denominator
First, we need to factor the quadratic expression in the denominator:
\[ x^2 + 5x + 6 = (x + 2)(x + 3) \]

### Step 2: Set Up Partial Fractions
We express the integrand as a sum of partial fractions:
\[ \frac{x + 1}{(x + 2)(x + 3)} = \frac{A}{x + 2} + \frac{B}{x + 3} \]
where \(A\) and \(B\) are constants to be determined.

### Step 3: Clear the Denominator
Multiply both sides by the denominator \((x + 2)(x + 3)\):
\[ x + 1 = A(x + 3) + B(x + 2) \]

### Step 4: Expand and Collect Terms
Expanding the right-hand side:
\[ x + 1 = Ax + 3A + Bx + 2B = (A + B)x + (3A + 2B) \]
This gives us two equations by comparing coefficients:
1. \(A + B = 1\) (coefficient of \(x\))
2. \(3A + 2B = 1\) (constant term)

### Step 5: Solve the System of Equations
From the first equation, we can express \(B\) in terms of \(A\):
\[ B = 1 - A \]
Substituting \(B\) into the second equation:
\[ 3A + 2(1 - A) = 1 \\ 3A + 2 - 2A = 1 \\ A + 2 = 1 \\ A = -1 \]
Now substituting \(A\) back to find \(B\):
\[ B = 1 - (-1) = 2 \]

### Step 6: Write the Partial Fraction Decomposition
Now we can write:
\[ \frac{x + 1}{(x + 2)(x + 3)} = \frac{-1}{x + 2} + \frac{2}{x + 3} \]

### Step 7: Integrate Each Term
Now we can integrate each term separately:
\[ \int \left( \frac{-1}{x + 2} + \frac{2}{x + 3} \right) \, dx = \int \frac{-1}{x + 2} \, dx + \int \frac{2}{x + 3} \, dx \]
This results in:
\[ = -\ln |x + 2| + 2\ln |x + 3| + C \]

### Step 8: Combine the Logarithms
Using the properties of logarithms:
\[ = \ln \left| \frac{(x + 3)^2}{x + 2} \right| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{x+1}{x^2+5x+6} \, dx = \ln \left| \frac{(x + 3)^2}{x + 2} \right| + C \]