The problem asks to decompose the rational expression

$\frac{5x^2 - x + 2}{(x^2 + 1)(x - 1)}$

into partial fractions.

Step-by-step decomposition:

We need to express the given rational function as a sum of simpler fractions. The denominators $(x^2 + 1)(x - 1)$ suggest two types of terms: one for the linear factor $(x - 1)$ and one for the irreducible quadratic factor $(x^2 + 1)$.

The general form for the partial fraction decomposition would be:

$\frac{5x^2 - x + 2}{(x^2 + 1)(x - 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$

Where $A$, $B$, and $C$ are constants to be determined.

Step 1: Set up the equation

Multiply both sides of the equation by $(x^2 + 1)(x - 1)$ to eliminate the denominators:

$5x^2 - x + 2 = A(x^2 + 1) + (Bx + C)(x - 1)$

Step 2: Expand the right-hand side

Now expand both terms on the right-hand side:

$A(x^2 + 1) = A x^2 + A$ $(Bx + C)(x - 1) = Bx(x - 1) + C(x - 1) = Bx^2 - Bx + Cx - C$

So the equation becomes:

$5x^2 - x + 2 = A x^2 + A + Bx^2 - Bx + Cx - C$

Step 3: Combine like terms

Now group the terms by powers of $x$:

$5x^2 - x + 2 = (A + B)x^2 + (-B + C)x + (A - C)$

Step 4: Set up a system of equations

Now, compare the coefficients of corresponding powers of $x$ on both sides:

• Coefficient of $x^2$: $A + B = 5$
• Coefficient of $x$: $-B + C = -1$
• Constant term: $A - C = 2$

Step 5: Solve the system of equations

1. From $A + B = 5$, we get $B = 5 - A$.
2. From $-B + C = -1$, substitute $B = 5 - A$ to get: $-(5 - A) + C = -1 \quad \Rightarrow \quad -5 + A + C = -1 \quad \Rightarrow \quad A + C = 4$
3. From $A - C = 2$, solve for $C$: $A + C = 4 \quad \text{and} \quad A - C = 2$ Add the two equations to eliminate $C$: $2A = 6 \quad \Rightarrow \quad A = 3$
4. Substitute $A = 3$ into $A + C = 4$: $3 + C = 4 \quad \Rightarrow \quad C = 1$
5. Finally, substitute $A = 3$ into $B = 5 - A$: $B = 5 - 3 = 2$

Step 6: Write the partial fraction decomposition

Now that we have $A = 3$, $B = 2$, and $C = 1$, the partial fraction decomposition is:

$\frac{5x^2 - x + 2}{(x^2 + 1)(x - 1)} = \frac{3}{x - 1} + \frac{2x + 1}{x^2 + 1}$

Would you like further clarification or details?

Here are some related questions you might want to explore:

1. How do you handle complex roots in partial fraction decomposition?
2. What if the denominator had repeated factors?
3. Can partial fractions be applied to improper rational functions?
4. How can partial fractions help solve integrals?
5. Why is the decomposition important in signal processing?

Tip: Always check that the degree of the numerator is less than the denominator before decomposing into partial fractions; if not, perform polynomial division first!