Unit 8, lesson 64

Partial Fractions as an Algebraic Identity

How to split a proper rational expression into simpler fractions with linear and irreducible quadratic denominators, find the unknown numerators, and check the identity on its domain.

  • Theory part 127
  • Practice part 128
  • Rewriting Rational Functions

Adding fractions is a routine skill. You know that 2x−1+3x+2\frac{2}{x - 1} + \frac{3}{x + 2} combines into 5x+1(x−1)(x+2)\frac{5x + 1}{(x - 1)(x + 2)}. Partial fractions run that computation backwards, starting from the combined fraction and recovering the simple pieces. Calculus wants the pieces, because each one is easy to integrate and, in series, neighboring pieces often cancel.

Key terms

Proper rational expression

Definition A rational expression P(x)D(x)\frac{P(x)}{D(x)}, with PP and DD polynomials and DD not the zero polynomial, is proper when PP is zero or the degree of PP is less than the degree of DD. Partial fraction decomposition applies only to proper expressions. If the expression is improper, first divide to get P=DQ+RP = DQ + R with RR zero or of lower degree than DD, and then decompose the proper part RD\frac{R}{D}.

In plain words The top must have smaller degree than the bottom before you start. If it does not, long division comes first.

Example The expression x3x2−1\frac{x^3}{x^2 - 1} is improper. Division gives x3=(x2−1)x+xx^3 = (x^2 - 1)x + x, so

x3x2−1=x+xx2−1,x≠±1.\frac{x^3}{x^2 - 1} = x + \frac{x}{x^2 - 1}, \qquad x \ne \pm 1.

Only the proper part xx2−1\frac{x}{x^2 - 1} is split further. It equals 1/2x−1+1/2x+1\frac{1/2}{x - 1} + \frac{1/2}{x + 1}, as the next sections show how to find.

Common denominator

Definition To add rational expressions, rewrite each over a common denominator, usually the product of the distinct factors to their highest powers, and add the numerators. For nonzero polynomials BB and DD,

AB+CD=AD+CBBD,\frac{A}{B} + \frac{C}{D} = \frac{AD + CB}{BD},

valid wherever BB and DD are both nonzero. Recombining over a common denominator is how every partial fraction result is verified.

In plain words Bring the pieces back to one fraction. If you get the original expression, the decomposition is right.

Example Over the common denominator (x−1)(x+2)(x - 1)(x + 2),

2x−1+3x+2=2(x+2)+3(x−1)(x−1)(x+2)=5x+1(x−1)(x+2).\begin{aligned} &\frac{2}{x - 1} + \frac{3}{x + 2} \\[0.6em] &\quad = \frac{2(x + 2) + 3(x - 1)}{(x - 1)(x + 2)} \\[0.6em] &\quad = \frac{5x + 1}{(x - 1)(x + 2)}. \end{aligned}

Linear factors

Definition A linear factor of the denominator is a factor ax+bax + b with real a≠0a \ne 0. In a decomposition, a linear factor that appears once contributes one term Aax+b\frac{A}{ax + b} with a constant numerator AA. A linear factor that appears to the power kk contributes kk terms, one for each power,

A1ax+b+A2(ax+b)2+⋯+Ak(ax+b)k.\begin{aligned} &\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} \\[0.6em] &\quad + \cdots + \frac{A_k}{(ax + b)^k}. \end{aligned}

In plain words Every linear factor gets a constant on top. A repeated factor gets a separate term for each of its powers, not just the highest one.

Example For the repeated factor in x+3(x−1)2\frac{x + 3}{(x - 1)^2} use

x+3(x−1)2=Ax−1+B(x−1)2.\frac{x + 3}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}.

Multiplying by (x−1)2(x - 1)^2 gives x+3=A(x−1)+Bx + 3 = A(x - 1) + B. Matching the xx terms gives A=1A = 1, and then x=1x = 1 gives B=4B = 4. So the expression equals 1x−1+4(x−1)2\frac{1}{x - 1} + \frac{4}{(x - 1)^2} for x≠1x \ne 1.

Irreducible quadratic factors

Definition A quadratic ax2+bx+cax^2 + bx + c with real coefficients and a≠0a \ne 0 is irreducible over the reals when it has no real roots, that is when b2−4ac<0b^2 - 4ac < 0. Such a factor cannot be split into real linear factors. In a decomposition, an irreducible quadratic that appears once contributes one term with a numerator of degree at most one,

Bx+Cax2+bx+c.\frac{Bx + C}{ax^2 + bx + c}.

A repeated irreducible quadratic contributes one such term for each power, as with linear factors.

In plain words A quadratic that will not factor stays whole, and it gets a linear numerator, not just a constant, because a constant alone is not flexible enough.

Example The quadratic x2+1x^2 + 1 has discriminant 0−4=−4<00 - 4 = -4 < 0, so it is irreducible. The expression 3x2+x(x+1)(x2+1)\frac{3x^2 + x}{(x + 1)(x^2 + 1)} therefore takes the form

3x2+x(x+1)(x2+1)=Ax+1+Bx+Cx2+1.\begin{aligned} \frac{3x^2 + x}{(x + 1)(x^2 + 1)} &= \frac{A}{x + 1} \\[0.6em] &\quad + \frac{Bx + C}{x^2 + 1}. \end{aligned}

By contrast, x2−4x^2 - 4 has discriminant 16>016 > 0 and splits as (x−2)(x+2)(x - 2)(x + 2), so it must be factored before choosing the form.

Unknown coefficients and coefficient matching

Definition The numerators AA, BB, CC and so on in the chosen form are unknown coefficients. Multiplying both sides by the original denominator gives an equation between two polynomials. Coefficient matching sets the coefficients of each power of xx equal on both sides, which gives a system of linear equations for the unknowns. Substituting convenient values of xx, such as the roots of linear factors, gives equations for the same unknowns and is often faster.

In plain words Clear the denominators, then compare the two sides power by power. Each power gives one equation.

Example For the form above, multiply by (x+1)(x2+1)(x + 1)(x^2 + 1),

3x2+x=A(x2+1)+(Bx+C)(x+1).\begin{aligned} 3x^2 + x &= A(x^2 + 1) \\[0.4em] &\quad + (Bx + C)(x + 1). \end{aligned}

The right side expands to (A+B)x2+(B+C)x+(A+C)(A + B)x^2 + (B + C)x + (A + C). Matching powers gives three equations.

PowerEquation
x2x^2A+B=3A + B = 3
xxB+C=1B + C = 1
constantA+C=0A + C = 0

Adding the first and third and subtracting the second gives 2A=22A = 2, so A=1A = 1, then B=2B = 2 and C=−1C = -1. Setting x=−1x = -1 is a quick cross check, since it gives 3−1=2A3 - 1 = 2A, so A=1A = 1 again. Hence

3x2+x(x+1)(x2+1)=1x+1+2x−1x2+1,x≠−1.\begin{aligned} \frac{3x^2 + x}{(x + 1)(x^2 + 1)} &= \frac{1}{x + 1} \\[0.6em] &\quad + \frac{2x - 1}{x^2 + 1}, \\[0.6em] &\qquad x \ne -1. \end{aligned}

Partial fraction decomposition

Definition Let P(x)D(x)\frac{P(x)}{D(x)} be a proper rational expression with real coefficients, and factor DD into real linear factors and irreducible quadratic factors. Its partial fraction decomposition writes it as a sum of terms, one term A(ax+b)j\frac{A}{(ax + b)^j} with a real constant AA for each power jj up to the multiplicity of each linear factor and one term Bx+C(ax2+bx+c)j\frac{Bx + C}{(ax^2 + bx + c)^j} with real constants BB and CC for each power jj up to the multiplicity of each irreducible quadratic factor. Each term has its own unknown constants. Such a decomposition always exists and its coefficients are unique. Each term is called a partial fraction.

In plain words One fraction with a complicated denominator becomes a sum of fractions whose denominators are the simple pieces of the original.

Example Split 5x+1(x−1)(x+2)\frac{5x + 1}{(x - 1)(x + 2)} into Ax−1+Bx+2\frac{A}{x - 1} + \frac{B}{x + 2}. Clearing denominators gives

5x+1=A(x+2)+B(x−1).5x + 1 = A(x + 2) + B(x - 1).

At x=1x = 1 this reads 6=3A6 = 3A, so A=2A = 2. At x=−2x = -2 it reads −9=−3B-9 = -3B, so B=3B = 3. Matching coefficients agrees, since A+B=5A + B = 5 and 2A−B=12A - B = 1. The result is 2x−1+3x+2\frac{2}{x - 1} + \frac{3}{x + 2}, which the common denominator section already confirmed.

In series the same idea gives the telescoping identity

1k(k+1)=1k−1k+1,k≠0,−1.\begin{gathered} \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}, \\[0.6em] k \ne 0, -1. \end{gathered}

Domain restriction

Definition A partial fraction decomposition is an equality of fractions that holds exactly where the original denominator D(x)D(x) is nonzero. Multiplying both sides by D(x)D(x) gives an equation between two polynomials that is true at every real xx except the finitely many roots of DD. The difference of the two sides is then a polynomial with infinitely many roots, and the only such polynomial is the zero polynomial. So the cleared equation is a polynomial identity, true for every real xx, roots of DD included. That is why substituting a root of the denominator into the cleared equation is legitimate, even though the fractions themselves are undefined there.

In plain words The fractions live only where the denominator is not zero. The cleared polynomial equation is true everywhere, so you may plug the forbidden values into that equation, but never into the fractions.

Example In the decomposition of 5x+1(x−1)(x+2)\frac{5x + 1}{(x - 1)(x + 2)} the fractions are undefined at x=1x = 1 and x=−2x = -2. Substituting x=1x = 1 into 5x+1=A(x+2)+B(x−1)5x + 1 = A(x + 2) + B(x - 1) is fine, because that equation between polynomials holds at every real xx. A check at an allowed value such as x=0x = 0 gives −12-\frac{1}{2} on the left and 2−1+32=−12\frac{2}{-1} + \frac{3}{2} = -\frac{1}{2} on the right.

Check yourself

  1. Decompose 1x2−4\frac{1}{x^2 - 4} into partial fractions and state where the equality holds.
Show the answer

Factor x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2) and write 1=A(x+2)+B(x−2)1 = A(x + 2) + B(x - 2). Setting x=2x = 2 gives A=14A = \frac{1}{4} and setting x=−2x = -2 gives B=−14B = -\frac{1}{4}, so

1x2−4=1/4x−2−1/4x+2,x≠±2.\begin{gathered} \frac{1}{x^2 - 4} = \frac{1/4}{x - 2} - \frac{1/4}{x + 2}, \\[0.6em] x \ne \pm 2. \end{gathered}
  1. Write the form of the decomposition of x2+2(x−3)2(x2+x+1)\frac{x^2 + 2}{(x - 3)^2 (x^2 + x + 1)} without solving for the coefficients.
Show the answer

The quadratic x2+x+1x^2 + x + 1 has discriminant 1−4=−3<01 - 4 = -3 < 0, so it is irreducible. The repeated linear factor needs a term for each power, so the form is

Ax−3+B(x−3)2+Cx+Dx2+x+1.\frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{Cx + D}{x^2 + x + 1}.
  1. Decompose 3x2+2x−1(x−1)(x2+1)\frac{3x^2 + 2x - 1}{(x - 1)(x^2 + 1)} and verify by recombining.
Show the answer

Write 3x2+2x−1=A(x2+1)+(Bx+C)(x−1)3x^2 + 2x - 1 = A(x^2 + 1) + (Bx + C)(x - 1). Setting x=1x = 1 gives 4=2A4 = 2A, so A=2A = 2. Matching the x2x^2 terms gives A+B=3A + B = 3, so B=1B = 1, and matching constants gives A−C=−1A - C = -1, so C=3C = 3. Thus

3x2+2x−1(x−1)(x2+1)=2x−1+x+3x2+1,x≠1.\begin{aligned} \frac{3x^2 + 2x - 1}{(x - 1)(x^2 + 1)} &= \frac{2}{x - 1} \\[0.6em] &\quad + \frac{x + 3}{x^2 + 1}, \\[0.6em] &\qquad x \ne 1. \end{aligned}

Recombining, 2(x2+1)+(x+3)(x−1)=3x2+2x−12(x^2 + 1) + (x + 3)(x - 1) = 3x^2 + 2x - 1, which is the original numerator.

How this lesson connects