If `int(5x+2)/(x^(2)-3x+2)dx=log[(x-2)^(m).(x-1)^(n)]+c` then `(m,n)-=`

Options

A. `(12, -7)`
B. `(-12, 7)`
C. `(12, 7)`
D. `(-7, 12)`

Answer

To solve the integral

\[ \int \frac{5x + 2}{x^2 - 3x + 2} \, dx = \log[(x - 2)^m (x - 1)^n] + C, \]

we need to find the values of \(m\) and \(n\).

### Step 1: Factor the denominator
First, we factor the denominator \(x^2 - 3x + 2\):

\[ x^2 - 3x + 2 = (x - 2)(x - 1). \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:

\[ \int \frac{5x + 2}{(x - 2)(x - 1)} \, dx. \]

### Step 3: Perform partial fraction decomposition
We express the integrand as a sum of partial fractions:

\[ \frac{5x + 2}{(x - 2)(x - 1)} = \frac{A}{x - 2} + \frac{B}{x - 1}. \]

Multiplying through by the denominator \((x - 2)(x - 1)\) gives:

\[ 5x + 2 = A(x - 1) + B(x - 2). \]

### Step 4: Solve for \(A\) and \(B\)
Expanding the right-hand side:

\[ 5x + 2 = Ax - A + Bx - 2B = (A + B)x + (-A - 2B). \]

Equating coefficients, we have:

1. \(A + B = 5\) (coefficient of \(x\))
2. \(-A - 2B = 2\) (constant term)

From the first equation, we can express \(A\) in terms of \(B\):

\[ A = 5 - B. \]

Substituting into the second equation:

\[ -(5 - B) - 2B = 2 \implies -5 + B - 2B = 2 \implies -5 - B = 2 \implies B = -7. \]

Now substituting \(B\) back to find \(A\):

\[ A = 5 - (-7) = 12. \]

### Step 5: Rewrite the integral
Now we can rewrite the integral:

\[ \int \left( \frac{12}{x - 2} - \frac{7}{x - 1} \right) \, dx. \]

### Step 6: Integrate
Integrating term by term:

\[ \int \frac{12}{x - 2} \, dx - \int \frac{7}{x - 1} \, dx = 12 \log|x - 2| - 7 \log|x - 1| + C. \]

### Step 7: Combine logarithms
Using properties of logarithms, we can combine the logs:

\[ = \log \left( |(x - 2)^{12} (x - 1)^{-7}| \right) + C. \]

### Step 8: Identify \(m\) and \(n\)
From the expression \(\log[(x - 2)^m (x - 1)^n]\), we see that:

- \(m = 12\)
- \(n = -7\)

### Final Answer
Thus, the values of \(m\) and \(n\) are:

\[ (m, n) = (12, -7). \]