`int((2x+9))/((x+2)(x-3)^(2))dx`

Answer

To solve the integral \(\int \frac{2x + 9}{(x + 2)(x - 3)^2} \, dx\), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Set up the partial fraction decomposition

We want to express \(\frac{2x + 9}{(x + 2)(x - 3)^2}\) in the form:

\[ \frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2} \]

where \(A\), \(B\), and \(C\) are constants to be determined.

### Step 2: Combine the fractions

We can combine the right-hand side over a common denominator:

\[ \frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2} = \frac{A(x - 3)^2 + B(x + 2)(x - 3) + C(x + 2)}{(x + 2)(x - 3)^2} \]

### Step 3: Set the numerators equal

Now, we set the numerators equal to each other:

\[ 2x + 9 = A(x - 3)^2 + B(x + 2)(x - 3) + C(x + 2) \]

### Step 4: Expand the right-hand side

Expanding the right-hand side:

1. \(A(x - 3)^2 = A(x^2 - 6x + 9)\)
2. \(B(x + 2)(x - 3) = B(x^2 - x - 6)\)
3. \(C(x + 2) = Cx + 2C\)

Combining these gives:

\[ Ax^2 - 6Ax + 9A + Bx^2 - Bx - 6B + Cx + 2C \]

Combining like terms:

\[ (A + B)x^2 + (-6A - B + C)x + (9A - 6B + 2C) \]

### Step 5: Set up a system of equations

Now we equate coefficients from both sides:

1. Coefficient of \(x^2\): \(A + B = 0\)
2. Coefficient of \(x\): \(-6A - B + C = 2\)
3. Constant term: \(9A - 6B + 2C = 9\)

### Step 6: Solve the system of equations

From \(A + B = 0\), we can express \(B\) in terms of \(A\):

\[ B = -A \]

Substituting \(B = -A\) into the other equations:

1. \(-6A - (-A) + C = 2 \Rightarrow -5A + C = 2\)
2. \(9A - 6(-A) + 2C = 9 \Rightarrow 15A + 2C = 9\)

Now we have:

1. \(C = 5A + 2\)
2. Substituting into \(15A + 2(5A + 2) = 9\):

\[ 15A + 10A + 4 = 9 \Rightarrow 25A = 5 \Rightarrow A = \frac{1}{5} \]

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

\[ B = -\frac{1}{5}, \quad C = 5\left(\frac{1}{5}\right) + 2 = 3 \]

### Step 7: Write the partial fraction decomposition

Thus, we have:

\[ \frac{2x + 9}{(x + 2)(x - 3)^2} = \frac{1/5}{x + 2} - \frac{1/5}{x - 3} + \frac{3}{(x - 3)^2} \]

### Step 8: Integrate each term

Now we can integrate each term separately:

\[ \int \left( \frac{1/5}{x + 2} - \frac{1/5}{x - 3} + \frac{3}{(x - 3)^2} \right) dx \]

1. \(\int \frac{1/5}{x + 2} \, dx = \frac{1}{5} \ln |x + 2|\)
2. \(\int -\frac{1/5}{x - 3} \, dx = -\frac{1}{5} \ln |x - 3|\)
3. \(\int \frac{3}{(x - 3)^2} \, dx = -\frac{3}{x - 3}\)

### Step 9: Combine the results

Combining these results gives:

\[ \int \frac{2x + 9}{(x + 2)(x - 3)^2} \, dx = \frac{1}{5} \ln |x + 2| - \frac{1}{5} \ln |x - 3| - \frac{3}{x - 3} + C \]

### Final Answer

Thus, the final answer is:

\[ \frac{1}{5} \ln |x + 2| - \frac{1}{5} \ln |x - 3| - \frac{3}{x - 3} + C \]