To evaluate the integral
\[
\int \frac{x}{(x - 1)(x^2 + 1)} \, dx,
\]
we can use partial fraction decomposition to simplify the integrand.

1. **Set up the partial fraction decomposition**:

   We want to express the integrand as:
   \[
   \frac{x}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}
   \]
   where \( A \), \( B \), and \( C \) are constants to be determined.

2. **Combine terms**:
   Multiply both sides by the denominator \((x - 1)(x^2 + 1)\) to clear the fractions:
   \[
   x = A(x^2 + 1) + (Bx + C)(x - 1).
   \]

3. **Expand and equate coefficients**:
   Expanding the right-hand side, we get:
   \[
   x = A x^2 + A + Bx^2 - Bx + Cx - C.
   \]
   Combine like terms:
   \[
   x = (A + B)x^2 + (-B + C)x + (A - C).
   \]
   Now, equate coefficients of \(x^2\), \(x\), and the constant term on both sides:
   - For \(x^2\): \( A + B = 0 \)
   - For \(x\): \( -B + C = 1 \)
   - For the constant term: \( A - C = 0 \)

4. **Solve the system of equations**:
   From \( A + B = 0 \), we have \( B = -A \).
   Substitute \( B = -A \) into the second equation:
   \[
   -(-A) + C = 1 \Rightarrow A + C = 1.
   \]
   From the third equation \( A - C = 0 \), we get \( C = A \).

   Substitute \( C = A \) into \( A + C = 1 \):
   \[
   A + A = 1 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2}.
   \]
   Then \( B = -\frac{1}{2} \) and \( C = \frac{1}{2} \).

5. **Rewrite the integrand**:
   Substitute \( A \), \( B \), and \( C \) back into the partial fraction decomposition:
   \[
   \frac{x}{(x - 1)(x^2 + 1)} = \frac{1/2}{x - 1} + \frac{(-1/2)x + 1/2}{x^2 + 1}.
   \]
   This can be simplified as:
   \[
   = \frac{1}{2} \cdot \frac{1}{x - 1} + \frac{-x + 1}{2(x^2 + 1)}.
   \]

6. **Integrate each term**:
   Now we integrate each term separately:
   \[
   \int \frac{1}{2} \cdot \frac{1}{x - 1} \, dx + \int \frac{-x + 1}{2(x^2 + 1)} \, dx.
   \]

   - For the first integral:
     \[
     \int \frac{1}{2} \cdot \frac{1}{x - 1} \, dx = \frac{1}{2} \ln |x - 1|.
     \]

   - For the second integral, split it as follows:
     \[
     \int \frac{-x}{2(x^2 + 1)} \, dx + \int \frac{1}{2(x^2 + 1)} \, dx.
     \]
     The first part can be solved with the substitution \( u = x^2 + 1 \), \( du = 2x \, dx \):
     \[
     \int \frac{-x}{2(x^2 + 1)} \, dx = -\frac{1}{2} \int \frac{1}{u} \, du = -\frac{1}{2} \ln |x^2 + 1|.
     \]
     The second part is a standard arctangent integral:
     \[
     \int \frac{1}{2(x^2 + 1)} \, dx = \frac{1}{2} \arctan(x).
     \]

7. **Combine results**:
   Therefore, the integral is:
   \[
   \int \frac{x}{(x - 1)(x^2 + 1)} \, dx = \frac{1}{2} \ln |x - 1| - \frac{1}{2} \ln |x^2 + 1| + \frac{1}{2} \arctan(x) + C,
   \]
   where \( C \) is the constant of integration.

Would you like further details or have any questions? 

Here are some related questions:
1. How can substitution be used in integrals?
2. What are some examples of partial fraction decomposition?
3. Why is arctangent involved in integrals with quadratic denominators?
4. How does partial fraction decomposition help in rational functions?
5. What are common approaches to solve logarithmic integrals?

**Tip:** When you see rational functions with quadratic denominators, try partial fraction decomposition and look for arctangent in integrals.

Discover the step-by-step process to solve the integral of x / ((x - 1)(x^2 + 1)) using partial fraction decomposition. This solution covers calculus concepts, including logarithmic and arctangent integration, and is suitable for undergraduate calculus students.

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