`int(x^2+1)/(x(x^2-1))dx` is equal to

Options

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

Answer

To solve the integral \(\int \frac{x^2 + 1}{x(x^2 - 1)} \, dx\), we will use the method of partial fractions. Here’s a step-by-step breakdown of the solution:

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

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

### Step 3: Expand the right-hand side
Expanding the right-hand side:
\[ A(x^2 - 1) + B(x^2 + x) + C(x^2 - x) = (A + B + C)x^2 + (B - C)x + (-A) \]

### Step 4: Set up the system of equations
Now, we equate the coefficients from both sides:
1. For \(x^2\): \(A + B + C = 1\)
2. For \(x\): \(B - C = 0\)
3. Constant term: \(-A = 1\)

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

Thus, we have:
\[ A = -1, \quad B = 1, \quad C = 1 \]

### Step 6: Rewrite the integral
Now we can rewrite the integral:
\[ \int \left( \frac{-1}{x} + \frac{1}{x - 1} + \frac{1}{x + 1} \right) \, dx \]

### Step 7: Integrate term by term
Now we integrate each term separately:
\[ \int \frac{-1}{x} \, dx + \int \frac{1}{x - 1} \, dx + \int \frac{1}{x + 1} \, dx \]
This results in:
\[ - \ln |x| + \ln |x - 1| + \ln |x + 1| + C \]

### Step 8: Combine the logarithms
Using the properties of logarithms:
\[ \ln |x - 1| + \ln |x + 1| - \ln |x| = \ln \left( \frac{(x - 1)(x + 1)}{x} \right) + C \]
This simplifies to:
\[ \ln \left( \frac{x^2 - 1}{x} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{x^2 + 1}{x(x^2 - 1)} \, dx = \ln \left( \frac{x^2 - 1}{x} \right) + C \]