The partial fractions of `((x^(2)-1))/(x(x^(2)+1))` are

Options

A. `1/x + (2x)/(x^(2)+1)`
B. `- (1)/(x)+(x)/(x^(2)+1)`
C. `1/x -(2x)/(x^(2)+1)`
D. `- 1/x + (2x)/(x^(2)+1)`

Answer

To find the partial fractions of the expression \(\frac{x^2 - 1}{x(x^2 + 1)}\), we can follow these steps:

### Step 1: Set up the partial fraction decomposition
We start by expressing the given fraction as a sum of simpler fractions:
\[ \frac{x^2 - 1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \]
Here, \(A\), \(B\), and \(C\) are constants that we need to determine.

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

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

### Step 4: Compare coefficients
Now, we will compare the coefficients of both sides of the equation:
1. Coefficient of \(x^2\): \(A + B = 1\)
2. Coefficient of \(x\): \(C = 0\)
3. Constant term: \(A = -1\)

### Step 5: Solve the system of equations
From the constant term, we have:
\[ A = -1 \]
Substituting \(A = -1\) into the first equation:
\[ -1 + B = 1 \implies B = 2 \]
From the second equation, we already have:
\[ C = 0 \]

### Step 6: Write the partial fractions
Now substituting \(A\), \(B\), and \(C\) back into the partial fraction decomposition:
\[ \frac{x^2 - 1}{x(x^2 + 1)} = \frac{-1}{x} + \frac{2x}{x^2 + 1} \]

### Final Answer
Thus, the partial fractions of \(\frac{x^2 - 1}{x(x^2 + 1)}\) are:
\[ \frac{-1}{x} + \frac{2x}{x^2 + 1} \]