Resolve `(1)/((x+1)(x+2))` into partial fractions.

Answer

To resolve the expression \(\frac{1}{(x+1)(x+2)}\) into partial fractions, we can follow these steps:

### Step 1: Set up the partial fraction decomposition
We want to express \(\frac{1}{(x+1)(x+2)}\) in the form:
\[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \]
where \(A\) and \(B\) are constants that we need to determine.

### Step 2: Clear the denominators
To eliminate the denominators, we multiply both sides of the equation by \((x+1)(x+2)\):
\[ 1 = A(x+2) + B(x+1) \]

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

### Step 4: Set up equations for coefficients
Since the left side is a constant (1), we can equate the coefficients of \(x\) and the constant terms:
1. For the coefficient of \(x\): \(A + B = 0\)
2. For the constant term: \(2A + B = 1\)

### Step 5: Solve the system of equations
From the first equation \(A + B = 0\), we can express \(B\) in terms of \(A\):
\[ B = -A \]
Substituting \(B = -A\) into the second equation:
\[ 2A - A = 1 \implies A = 1 \]
Now substituting \(A = 1\) back into \(B = -A\):
\[ B = -1 \]

### Step 6: Write the partial fraction decomposition
Now that we have \(A\) and \(B\), we can write the partial fraction decomposition:
\[ \frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2} \]

### Final Result
Thus, the expression \(\frac{1}{(x+1)(x+2)}\) can be resolved into partial fractions as:
\[ \frac{1}{x+1} - \frac{1}{x+2} \]

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