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

Answer

To solve the integral \( \int \frac{x-1}{(x-1)^2(2x+3)} \, dx \), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Simplify the Integrand
First, we can simplify the integrand:
\[ \frac{x-1}{(x-1)^2(2x+3)} = \frac{1}{(x-1)(2x+3)} \]
This is because \( (x-1) \) in the numerator cancels with one \( (x-1) \) in the denominator.

### Step 2: Set Up Partial Fractions
Next, we express \( \frac{1}{(x-1)(2x+3)} \) in terms of partial fractions:
\[ \frac{1}{(x-1)(2x+3)} = \frac{A}{x-1} + \frac{B}{2x+3} \]
where \( A \) and \( B \) are constants we need to determine.

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

### Step 4: Expand and Collect Like Terms
Expanding the right side:
\[ 1 = 2Ax + 3A + Bx - B \]
Combining like terms gives:
\[ 1 = (2A + B)x + (3A - B) \]

### Step 5: Set Up the System of Equations
Now, we can equate the coefficients from both sides:
1. \( 2A + B = 0 \) (coefficient of \( x \))
2. \( 3A - B = 1 \) (constant term)

### Step 6: Solve the System of Equations
From the first equation, we can express \( B \) in terms of \( A \):
\[ B = -2A \]
Substituting \( B \) into the second equation:
\[ 3A - (-2A) = 1 \implies 3A + 2A = 1 \implies 5A = 1 \implies A = \frac{1}{5} \]
Now substituting \( A \) back to find \( B \):
\[ B = -2 \left(\frac{1}{5}\right) = -\frac{2}{5} \]

### Step 7: Rewrite the Integral
Now we can rewrite the integral using the values of \( A \) and \( B \):
\[ \int \left( \frac{1/5}{x-1} - \frac{2/5}{2x+3} \right) \, dx \]

### Step 8: Integrate Each Term
Now we can integrate each term separately:
\[ \int \frac{1/5}{x-1} \, dx - \int \frac{2/5}{2x+3} \, dx \]
The integrals are:
\[ \frac{1}{5} \ln |x-1| - \frac{2}{5} \cdot \frac{1}{2} \ln |2x+3| + C \]
This simplifies to:
\[ \frac{1}{5} \ln |x-1| - \frac{1}{5} \ln |2x+3| + C \]

### Step 9: Combine the Logarithms
Using properties of logarithms:
\[ \frac{1}{5} \left( \ln |x-1| - \ln |2x+3| \right) + C = \frac{1}{5} \ln \left| \frac{x-1}{2x+3} \right| + C \]

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