Resolve `(2x+7)/((x+1)(x^(2)+4))` into partial fractions.

Answer

To resolve the expression \(\frac{2x + 7}{(x + 1)(x^2 + 4)}\) into partial fractions, we follow these steps:

### Step 1: Set up the partial fraction decomposition
We express \(\frac{2x + 7}{(x + 1)(x^2 + 4)}\) as a sum of fractions:
\[ \frac{2x + 7}{(x + 1)(x^2 + 4)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 4} \]
where \(A\), \(B\), and \(C\) are constants to be determined.

### Step 2: Combine the right-hand side
We need a common denominator to combine the fractions on the right:
\[ \frac{A(x^2 + 4) + (Bx + C)(x + 1)}{(x + 1)(x^2 + 4)} = \frac{2x + 7}{(x + 1)(x^2 + 4)} \]

### Step 3: Expand the numerator
Now, we expand the numerator:
\[ A(x^2 + 4) + (Bx + C)(x + 1) = Ax^2 + 4A + Bx^2 + Bx + Cx + C \]
Combining like terms gives:
\[ (A + B)x^2 + (B + C)x + (4A + C) \]

### Step 4: Set up the equation
Since the denominators are equal, we equate the numerators:
\[ (A + B)x^2 + (B + C)x + (4A + C) = 2x + 7 \]

### Step 5: Compare coefficients
From the equation above, we can compare coefficients:
1. For \(x^2\): \(A + B = 0\)
2. For \(x\): \(B + C = 2\)
3. Constant term: \(4A + C = 7\)

### Step 6: Solve the system of equations
From the first equation, we can express \(A\) in terms of \(B\):
\[ A = -B \]

Substituting \(A = -B\) into the second equation:
\[ B + C = 2 \implies -A + C = 2 \implies C = 2 + A \]

Now substituting \(C = 2 + A\) into the third equation:
\[ 4A + (2 + A) = 7 \implies 5A + 2 = 7 \implies 5A = 5 \implies A = 1 \]

Now substituting \(A = 1\) back to find \(B\) and \(C\):
\[ B = -A = -1 \]
\[ C = 2 + A = 2 + 1 = 3 \]

### Step 7: Write the final partial fraction decomposition
Now we substitute \(A\), \(B\), and \(C\) back into the partial fractions:
\[ \frac{2x + 7}{(x + 1)(x^2 + 4)} = \frac{1}{x + 1} + \frac{-x + 3}{x^2 + 4} \]

### Final Answer:
\[ \frac{2x + 7}{(x + 1)(x^2 + 4)} = \frac{1}{x + 1} + \frac{-x + 3}{x^2 + 4} \]