`int1/((x-1)(x^(2)+1))dx=`

Options

A. `1/2log(x-1)-1/4log(x^(2)+1)-1/2tan^(-1)x+c`
B. `1/2log(x-1)+1/4log(x^(2)+1)-1/2tan^(-1)x+c`
C. `1/2log(x-1)-1/2log(x^(2)+1)-1/2tan^(-1)x+c`
D. None of these

Answer

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

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

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

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

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

### Step 5: Solve the System of Equations
From the first equation, we have:
\[ B = -A \]
Substituting into the second equation:
\[ C - (-A) = 0 \implies C = A \]
Now substituting \( C = A \) into the third equation:
\[ A - A = 1 \implies 0 = 1 \]
This is incorrect, so let's correct our approach.

Instead, we can solve the equations:
1. \( A + B = 0 \) implies \( B = -A \)
2. \( C = -B \) implies \( C = A \)
3. Substitute \( C = A \) into \( A - C = 1 \):
\[ A - A = 1 \implies 0 = 1 \]
This indicates we need to re-evaluate the coefficients.

### Step 6: Find Values for A, B, and C
Let’s substitute specific values for \( x \) to find \( A \), \( B \), and \( C \):
- Let \( x = 1 \):
\[ 1 = A(1^2 + 1) + (B(1) + C)(1 - 1) \implies 1 = 2A \implies A = \frac{1}{2} \]
- Now, \( B = -A = -\frac{1}{2} \)
- And \( C = A = \frac{1}{2} \)

### Step 7: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{1}{(x-1)(x^2+1)} \, dx = \int \left( \frac{1/2}{x-1} + \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2 + 1} \right) \, dx \]

### Step 8: Integrate Each Term
Now we integrate each term separately:
1. \( \int \frac{1/2}{x-1} \, dx = \frac{1}{2} \ln |x-1| + C_1 \)
2. \( \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2 + 1} \, dx = -\frac{1}{2} \int \frac{x}{x^2 + 1} \, dx + \frac{1}{2} \int \frac{1}{x^2 + 1} \, dx \)

The first integral can be solved using substitution \( u = x^2 + 1 \), and the second integral is \( \frac{1}{2} \tan^{-1}(x) + C_2 \).

### Final Result
Putting it all together, we have:
\[ \int \frac{1}{(x-1)(x^2+1)} \, dx = \frac{1}{2} \ln |x-1| - \frac{1}{4} \ln(x^2 + 1) + \frac{1}{2} \tan^{-1}(x) + C \]