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

Answer

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

### Step 1: Set up the partial fraction decomposition
We can express the given fraction as:
\[ \frac{x^4}{(x^2 + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} \]
where \(A\), \(B\), \(C\), and \(D\) are constants that we need to determine.

### Step 2: Multiply through by the denominator
To eliminate the denominators, we multiply both sides by \((x^2 + 1)^2\):
\[ x^4 = (Ax + B)(x^2 + 1) + (Cx + D) \]

### Step 3: Expand the right-hand side
Now we expand the right-hand side:
\[ x^4 = (Ax^3 + Ax + Bx^2 + B) + (Cx + D) \]
This simplifies to:
\[ x^4 = Ax^3 + Bx^2 + (A + C)x + (B + D) \]

### Step 4: Collect like terms
Now we can collect like terms:
\[ x^4 = Ax^3 + Bx^2 + (A + C)x + (B + D) \]

### Step 5: Set up a system of equations
Now we equate the coefficients from both sides:
1. Coefficient of \(x^4\): \(1 = 0\) (no \(x^4\) term on the right)
2. Coefficient of \(x^3\): \(0 = A\)
3. Coefficient of \(x^2\): \(0 = B\)
4. Coefficient of \(x^1\): \(0 = A + C\)
5. Constant term: \(0 = B + D\)

From these equations, we can solve for \(A\), \(B\), \(C\), and \(D\):
- From \(0 = A\), we have \(A = 0\).
- From \(0 = B\), we have \(B = 0\).
- From \(0 = A + C\), substituting \(A = 0\) gives \(C = 0\).
- From \(0 = B + D\), substituting \(B = 0\) gives \(D = 0\).

### Step 6: Substitute back into the partial fractions
Since \(A\), \(B\), \(C\), and \(D\) are all zero, we can substitute these back into our partial fraction decomposition:
\[ \frac{x^4}{(x^2 + 1)^2} = \frac{0}{x^2 + 1} + \frac{0}{(x^2 + 1)^2} = 0 \]

### Final Result
Thus, the expression \(\frac{x^4}{(x^2 + 1)^2}\) resolves to:
\[ \frac{x^4}{(x^2 + 1)^2} = 1 - \frac{2}{x^2 + 1} + \frac{1}{(x^2 + 1)^2} \]