Let's work on each problem step-by-step.

Problem 1

The expression is: $\frac{5x + 1}{x^3 - x}$

1. Factor the denominator: We can factor $x^3 - x$ as follows: $x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)$

2. Set up the partial fraction decomposition: Since the factors are $x$, $x - 1$, and $x + 1$, we can write: $\frac{5x + 1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$

3. Clear the denominators: Multiply both sides by $x(x - 1)(x + 1)$ to obtain: $5x + 1 = A(x - 1)(x + 1) + B(x)(x + 1) + C(x)(x - 1)$

4. Expand and solve for $A$, $B$, and $C$ by comparing coefficients.

Problem 2

The expression is: $\frac{x^3 - x^2 + 2x + 4}{x^2 (x - 2)^2}$

1. Factor the denominator: The denominator is already factored as $x^2 (x - 2)^2$.

2. Set up the partial fraction decomposition: Since we have a repeated factor $x$ and $(x - 2)$, we write: $\frac{x^3 - x^2 + 2x + 4}{x^2 (x - 2)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2} + \frac{D}{(x - 2)^2}$

3. Clear the denominators: Multiply both sides by $x^2 (x - 2)^2$ and then expand and compare coefficients to solve for $A$, $B$, $C$, and $D$.

Would you like detailed solutions for each of these steps, or do you have any questions?

Additional Questions for Deeper Understanding

1. How do we determine the form of partial fractions when we have repeated linear factors?
2. Why do we use different constants (like $A$, $B$, etc.) for each term in the decomposition?
3. What happens if a polynomial in the numerator has a higher degree than the denominator?
4. How can we check if our partial fraction decomposition is correct?
5. How does partial fraction decomposition apply in integral calculus?

Tip

When decomposing fractions, always start by factoring the denominator completely—this simplifies the setup of partial fractions.