6: Partial fraction decomposition

Partial Fraction Decomposition

Partial fraction decomposition (PFD) is an essential algebraic technique for integrating rational functions (ratios of polynomials) where the degree of the numerator is less than the degree of the denominator. If the numerator degree is equal to or higher, polynomial long division must be performed first to obtain a proper rational function plus a polynomial.

The core idea is to decompose a complex rational expression into a sum of simpler fractions whose denominators correspond to the factors of the original denominator. These simpler fractions are then much easier to integrate.

Steps for Decomposition:

1. Factor the Denominator: Completely factor the denominator into linear factors $(ax + b)$ and irreducible quadratic factors $(ax^2 + bx + c)$.
2. Set Up the Decomposition: Write the original fraction as a sum of partial fractions. Each factor in the denominator dictates the form of the terms:
   • Distinct Linear Factor $(ax + b)$: Contributes a term $\frac{A}{ax + b}$.
   • Repeated Linear Factor $(ax + b)^k$: Contributes terms $\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \dots + \frac{A_k}{(ax + b)^k}$.
   • Distinct Irreducible Quadratic Factor $(ax^2 + bx + c)$: Contributes a term $\frac{Ax + B}{ax^2 + bx + c}$.
   • Repeated Irreducible Quadratic Factor $(ax^2 + bx + c)^k$: Contributes terms $\frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \dots + \frac{A_kx + B_k}{(ax^2 + bx + c)^k}$.
3. Solve for Constants: Multiply both sides of the equation by the original denominator. Expand the right-hand side and collect like terms. Equate the coefficients of corresponding powers of $x$ on both sides to form a system of linear equations. Solve this system for the unknown constants $A$, $B$, $A_1$, $B_1$, etc.

Example: Decompose $\frac{3x + 5}{(x-1)(x+2)}$.

1. Factors are distinct linear: $(x-1)$, $(x+2)$.
2. Set up: $\frac{3x + 5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$.
3. Multiply both sides by $(x-1)(x+2)$: $3x + 5 = A(x+2) + B(x-1)$
4. Solve for $A$ and $B$:
   • Method 1 (Substitution): Choose values for $x$ that eliminate one constant.
     • Let $x = 1$: $3(1) + 5 = A(1+2) + B(0) \implies 8 = 3A \implies A = 8/3$.
     • Let $x = -2$: $3(-2) + 5 = A(0) + B(-2-1) \implies -1 = -3B \implies B = 1/3$.
   • Method 2 (Equating Coefficients): Expand the right side: $Ax + 2A + Bx - B = (A+B)x + (2A - B)$. Set equal to left side: $A + B = 3 \quad \text{(coefficients of } x\text{)}$ $2A - B = 5 \quad \text{(constant terms)}$ Solving this system also gives $A = 8/3$, $B = 1/3$.
5. Result: $\frac{3x + 5}{(x-1)(x+2)} = \frac{8/3}{x-1} + \frac{1/3}{x+2} = \frac{8}{3(x-1)} + \frac{1}{3(x+2)}$.

Integration: After decomposition, integrate each partial fraction term: $\int \frac{3x+5}{(x-1)(x+2)} dx = \int \frac{8}{3(x-1)} dx + \int \frac{1}{3(x+2)} dx = \frac{8}{3} \ln|x-1| + \frac{1}{3} \ln|x+2| + C$

Key Exam Tips:

• Check Degree: Always ensure the numerator degree is less than the denominator degree before decomposing.
• Irreducible Quadratics: Remember numerators for quadratic factors are linear ($Ax+B$), not constants.
• Repeated Factors: Include terms for every power up to the multiplicity.
• Solving System: Substitution (using roots) is often fastest for linear factors. Equating coefficients is more general and necessary for quadratics/repeated factors. Be meticulous with algebra.
• Integration: After PFD, integrals typically become $\int \frac{A}{ax+b} dx = \frac{A}{a}\ln|ax+b|$ or $\int \frac{Ax+B}{ax^2+bx+c} dx$ (often requiring completing the square and substitution).