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

Options

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

Answer

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

### Step 1: Factor the Denominator
The denominator can be factored as follows:
\[ x(x^2 - 1) = x(x + 1)(x - 1) \]
So, we rewrite the expression:
\[ \frac{x^2 + 1}{x(x + 1)(x - 1)} \]

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

### Step 3: Clear the Denominator
Multiply both sides by the denominator \(x(x + 1)(x - 1)\) to eliminate the fraction:
\[ x^2 + 1 = A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) \]

### Step 4: Expand the Right Side
Now, expand the right side:
1. \(A(x + 1)(x - 1) = A(x^2 - 1)\)
2. \(Bx(x - 1) = B(x^2 - x)\)
3. \(Cx(x + 1) = C(x^2 + x)\)

Combining these gives:
\[ x^2 + 1 = Ax^2 - A + Bx^2 - Bx + Cx^2 + Cx \]
Combining like terms results in:
\[ x^2 + 1 = (A + B + C)x^2 + (-B + C)x - A \]

### Step 5: Set Up the System of Equations
Now, equate the coefficients from both sides:
1. Coefficient of \(x^2\): \(A + B + C = 1\)
2. Coefficient of \(x\): \(-B + C = 0\)
3. Constant term: \(-A = 1\)

### Step 6: Solve the System of Equations
From the third equation, we find:
\[ A = -1 \]
Substituting \(A\) into the first equation:
\[ -1 + B + C = 1 \implies B + C = 2 \quad \text{(Equation 1)} \]
From the second equation:
\[ -B + C = 0 \implies C = B \quad \text{(Equation 2)} \]
Substituting Equation 2 into Equation 1:
\[ B + B = 2 \implies 2B = 2 \implies B = 1 \]
Then substituting back to find \(C\):
\[ C = B = 1 \]

### Step 7: Write the Partial Fraction Decomposition
Now we have:
\[ A = -1, \quad B = 1, \quad C = 1 \]
Thus, the partial fraction decomposition is:
\[ \frac{x^2 + 1}{x(x + 1)(x - 1)} = \frac{-1}{x} + \frac{1}{x + 1} + \frac{1}{x - 1} \]

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