Let's break down and solve the rational equation step-by-step:

Given equation:

$\frac{2}{x^2 + 1} - \frac{1}{x - 1} = \frac{1}{2}$

Step 1: Find the Least Common Denominator (LCD).

The denominators are $x^2 + 1$, $x - 1$, and 2. The LCD is simply the product of these distinct denominators:

$\text{LCD} = 2(x^2 + 1)(x - 1)$

Step 2: Multiply both sides of the equation by the LCD.

Multiply each term by the LCD to eliminate the fractions:

$2(x^2 + 1)(x - 1) \left(\frac{2}{x^2 + 1}\right) - 2(x^2 + 1)(x - 1) \left(\frac{1}{x - 1}\right) = 2(x^2 + 1)(x - 1) \left(\frac{1}{2}\right)$

This simplifies to:

$4(x - 1) - 2(x^2 + 1) = (x^2 + 1)(x - 1)$

Step 3: Solve the equation.

Expand and simplify each side:

Left side: $4x - 4 - 2x^2 - 2$ Which becomes: $-2x^2 + 4x - 6$

Right side: First, expand $(x^2 + 1)(x - 1)$: $(x^2 + 1)(x - 1) = x^3 - x^2 + x - 1$

Now the equation is: $-2x^2 + 4x - 6 = x^3 - x^2 + x - 1$

Bring all terms to one side: $0 = x^3 + x^2 - 3x + 5$

Step 4: Solve the cubic equation.

To solve this cubic equation, we would normally use methods like factoring, synthetic division, or numerical techniques. However, this specific cubic does not factor nicely, so we might need to rely on numerical approximation or graphing to find the roots.

Final Answer:

The exact roots of the equation are complex and not easily factored, so you would typically approximate them using numerical methods or a calculator.

Would you like more details on the steps or further explanation on how to approximate the roots?

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Here are five related questions for further practice:

1. Solve the rational equation: $\frac{3}{x + 2} - \frac{1}{x - 3} = \frac{2}{x^2 - x - 6}$.
2. What are the steps to determine extraneous solutions in a rational equation?
3. How can you solve rational inequalities using a sign chart?
4. Why is it important to find the LCD when solving rational equations?
5. What is the difference between solving rational equations and solving rational inequalities?

Tip: Always check your solutions in the original equation to ensure they don't make any denominator zero.