7.3 Partial Fractions Decomposition

Partial Fractions Decomposition

Partial fractions decomposition is a technique for breaking a complex rational expression into a sum of simpler fractions. This makes it possible to integrate rational functions and solve equations that would be extremely difficult to work with in their original form.

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Decomposition Requirements

Before you can decompose a rational expression, two conditions must be met:

• The degree of the numerator must be less than the degree of the denominator. If it isn't, you need to perform polynomial long division first, then decompose the remainder.
• The denominator must be factored completely into linear factors and/or irreducible quadratic factors (quadratics that can't be factored over the reals).

Once those conditions are satisfied, the setup depends on what kinds of factors appear in the denominator. There are three cases:

• Case 1: Distinct linear factors (e.g., $\frac{1}{(x-1)(x+2)}$)
• Case 2: Repeated linear factors (e.g., $\frac{1}{(x-1)^2(x+2)}$)
• Case 3: Irreducible quadratic factors (e.g., $\frac{1}{(x^2+1)(x+2)}$)

A single problem can involve a mix of these cases. The rules below tell you what to write for each type of factor.

Case 1: Distinct Linear Factors

Each distinct linear factor gets its own fraction with an unknown constant in the numerator.

Example: Decompose $\frac{2x+1}{(x-1)(x+2)}$

1. Write the decomposition template: $\frac{2x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

2. Multiply both sides by the common denominator $(x-1)(x+2)$: $2x+1 = A(x+2) + B(x-1)$

3. Solve for A and B. The fastest method is to plug in the values that make each factor zero:

   • Set $x = 1$: $2(1)+1 = A(3) + B(0)$ → $3 = 3A$ → $A = 1$
   • Set $x = -2$: $2(-2)+1 = A(0) + B(-3)$ → $-3 = -3B$ → $B = 1$

4. Write the result: $\frac{2x+1}{(x-1)(x+2)} = \frac{1}{x-1} + \frac{1}{x+2}$

Case 2: Repeated Linear Factors

When a linear factor is raised to a power, you need a separate fraction for every power of that factor, from 1 up to the highest power.

Example: Decompose $\frac{2x+1}{(x-1)^2(x+2)}$

1. Write the template (notice the two fractions for $(x-1)$): $\frac{2x+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$

2. Multiply both sides by $(x-1)^2(x+2)$: $2x+1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$

3. Use strategic substitution and/or expand and match coefficients to solve for A, B, and C.

   • Set $x = 1$: $3 = B(3)$ → $B = 1$
   • Set $x = -2$: $-3 = C(9)$ → $C = -\frac{1}{3}$
   • Set $x = 0$ (or any convenient value): $1 = A(-1)(2) + 1(2) + (-\frac{1}{3})(1)$ → $1 = -2A + 2 - \frac{1}{3}$ → $A = \frac{1}{3}$

Case 3: Irreducible Quadratic Factors

An irreducible quadratic factor (like $x^2+1$, where the discriminant is negative) requires a linear numerator of the form $Ax + B$, not just a single constant.

Example: Decompose $\frac{2x+1}{(x^2+1)(x+2)}$

1. Write the template: $\frac{2x+1}{(x^2+1)(x+2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+2}$

2. Multiply both sides by $(x^2+1)(x+2)$: $2x+1 = (Ax+B)(x+2) + C(x^2+1)$

3. Use substitution where possible, then expand and match coefficients of like powers of $x$ to solve for A, B, and C.

   • Set $x = -2$: $-3 = C(5)$ → $C = -\frac{3}{5}$
   • Expand the right side and group by powers of $x$, then match coefficients of $x^2$, $x$, and the constant term to find A and B.

Integration with Partial Fractions

Once a rational expression is decomposed, each piece can be integrated using techniques you already know.

Integration Results by Factor Type

Distinct linear factor: $\int \frac{A}{x-a}\, dx = A \ln|x-a| + C$

Repeated linear factor: $\int \frac{A}{(x-a)^n}\, dx = \frac{A}{-(n-1)(x-a)^{n-1}} + C \quad \text{for } n \geq 2$

For example: $\int \frac{A}{(x-1)^2}\, dx = -\frac{A}{x-1} + C$

Irreducible quadratic factor (for $x^2 + a^2$): $\int \frac{Ax+B}{x^2+a^2}\, dx = \frac{A}{2}\ln(x^2+a^2) + \frac{B}{a}\arctan\!\left(\frac{x}{a}\right) + C$

The $Ax$ part produces a logarithm (via u-substitution with $u = x^2 + a^2$), and the $B$ part produces an arctangent.

Putting It Together

To integrate a rational function using partial fractions:

1. Check that the numerator degree is less than the denominator degree. If not, do long division first.
2. Factor the denominator completely.
3. Decompose into partial fractions.
4. Integrate each fraction separately using the results above.
5. Combine the results and add the constant of integration.

Example: $\int \frac{2x+1}{(x-1)(x+2)}\, dx = \int \frac{1}{x-1}\, dx + \int \frac{1}{x+2}\, dx = \ln|x-1| + \ln|x+2| + C$

Solving Rational Equations with Partial Fractions

Partial fractions can also help solve rational equations by simplifying complex expressions into pieces that are easier to work with.

Process

1. Identify restrictions. Find the values of $x$ that make the denominator zero. These are excluded from the domain. For $\frac{2x+1}{(x-1)(x+2)} = 0$, you'd have $x \neq 1$ and $x \neq -2$.

2. Decompose the rational expression on one side of the equation using partial fractions.

3. Solve the simplified equation. With simpler fractions, you can multiply through by the common denominator and solve the resulting polynomial equation.

4. Check your solutions against the restrictions from Step 1. Any solution that makes a denominator zero is extraneous and must be thrown out.