`int (dx)/((x+1)(x-2))=A log (x+1)+B log (x-2)+C`, where

Options

A. `A+B=0`
B. `AB=0`
C. `A//B=-1`
D. None of these

Answer

To solve the integral \(\int \frac{dx}{(x+1)(x-2)}\) and express it in the form \(A \log(x+1) + B \log(x-2) + C\), we can use the method of partial fractions. Here’s the step-by-step solution:

### Step 1: Set up the partial fraction decomposition
We start by expressing the integrand as a sum of partial fractions:
\[ \frac{1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} \]
where \(A\) and \(B\) are constants to be determined.

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

### Step 3: Expand the right-hand side
Expanding the right-hand side gives:
\[ 1 = Ax - 2A + Bx + B \]
Combining like terms, we have:
\[ 1 = (A + B)x + (-2A + B) \]

### Step 4: Set up equations for coefficients
For the equation to hold for all \(x\), the coefficients of \(x\) and the constant terms must be equal. This gives us the system of equations:
1. \(A + B = 0\) (coefficient of \(x\))
2. \(-2A + B = 1\) (constant term)

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

### Step 6: Write the integral using the partial fractions
Now we can rewrite the integral:
\[ \int \frac{dx}{(x+1)(x-2)} = \int \left(-\frac{1}{3(x+1)} + \frac{1}{3(x-2)}\right) dx \]

### Step 7: Integrate each term
Integrating each term separately:
\[ = -\frac{1}{3} \int \frac{dx}{x+1} + \frac{1}{3} \int \frac{dx}{x-2} \]
\[ = -\frac{1}{3} \log|x+1| + \frac{1}{3} \log|x-2| + C \]

### Step 8: Combine the logarithms
Combining the logarithms gives:
\[ = \frac{1}{3} \left(\log|x-2| - \log|x+1|\right) + C = \frac{1}{3} \log\left|\frac{x-2}{x+1}\right| + C \]

### Conclusion
Thus, we have:
\[ \int \frac{dx}{(x+1)(x-2)} = -\frac{1}{3} \log(x+1) + \frac{1}{3} \log(x-2) + C \]

### Final Result
Identifying \(A = -\frac{1}{3}\) and \(B = \frac{1}{3}\), we can conclude:
\[ A + B = 0 \quad \text{and} \quad \frac{A}{B} = -1 \]