Resolve `(2x^(2)+2x+1)/((x^(3)+x^(2))` into partial fractions.

Answer

To resolve the expression \(\frac{2x^2 + 2x + 1}{x^3 + x^2}\) into partial fractions, we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \(x^3 + x^2\):
\[ x^3 + x^2 = x^2(x + 1) \]

### Step 2: Set Up the Partial Fraction Decomposition
Next, we express the fraction in terms of partial fractions:
\[ \frac{2x^2 + 2x + 1}{x^2(x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1} \]
where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 3: Combine the Right Side
Now, we combine the right side over a common denominator:
\[ \frac{A(x)(x + 1) + B(x + 1) + C(x^2)}{x^2(x + 1)} = \frac{2x^2 + 2x + 1}{x^2(x + 1)} \]
This gives us the equation:
\[ A(x^2 + x) + B(x + 1) + C(x^2) = 2x^2 + 2x + 1 \]

### Step 4: Expand and Collect Like Terms
Expanding the left side:
\[ Ax^2 + Ax + Bx + B + Cx^2 = (A + C)x^2 + (A + B)x + B \]
Now, we equate the coefficients from both sides:
1. For \(x^2\): \(A + C = 2\)
2. For \(x\): \(A + B = 2\)
3. For the constant term: \(B = 1\)

### Step 5: Solve the System of Equations
From the third equation, we have:
\[ B = 1 \]
Substituting \(B\) into the second equation:
\[ A + 1 = 2 \implies A = 1 \]
Now substituting \(A\) into the first equation:
\[ 1 + C = 2 \implies C = 1 \]

### Step 6: Write the Partial Fractions
Now we have found the values:
\[ A = 1, \quad B = 1, \quad C = 1 \]
Thus, we can write the partial fraction decomposition as:
\[ \frac{2x^2 + 2x + 1}{x^3 + x^2} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x + 1} \]

### Final Answer
The final result of the partial fraction decomposition is:
\[ \frac{2x^2 + 2x + 1}{x^3 + x^2} = \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x + 1} \]