Partial Fractions

Main idea

Adding fractions builds a common denominator. Partial fractions runs that backwards: once you factor the denominator, a complicated rational function splits into simple fractions, one per factor, and each of those integrates to a logarithm or an arctangent.

\frac{x - 1}{({\htmlData{sym=2}{x - 2}})(x - 3)} = \frac{\htmlData{sym=0}{{A}}}{{\htmlData{sym=2}{x - 2}}} + \frac{\htmlData{sym=1}{{B}}}{x - 3}

What the symbols mean

$A$
Unknown numerator
A constant you solve for. Plug in x = 2 to make the other term vanish.
$B$
Unknown numerator
Found the same way by plugging in x = 3.
$x - 2$
Linear factor
One factor of the denominator. Each one contributes a fraction.

Explanation

Partial fractions is simply the process of breaking one complicated fraction of polynomials into a sum of small, simple fractions. The name comes from those small fractions: each one is a part of the original, and they add back up to the whole. It's good for integrating any rational function, and it's how you solve the logistic growth equation and undo Laplace transforms later on. All we're really saying with $\frac{x - 1}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$ is: the left side (the fraction you're stuck with) equals one simple piece for each factor of the bottom, with $A$ and $B$ (plain numbers) still to be found, and once you have them each piece integrates to a logarithm. The table below looks like a lot, but it's only bookkeeping for which pieces to write; bear with it.

Think back to adding fractions in algebra:

\frac{-1}{x - 2} + \frac{2}{x - 3} = \frac{-(x - 3) + 2(x - 2)}{(x - 2)(x - 3)} = \frac{x - 1}{(x - 2)(x - 3)}.

Going left to right is easy. But for integrating, you want to go right to left. $\int \frac{x - 1}{(x-2)(x-3)}\,dx$ looks hard, while $\int \frac{-1}{x - 2}\,dx$ and $\int \frac{2}{x - 3}\,dx$ are just logarithms. So the question is: given the combined fraction, how do you recover the pieces?

The denominator tells you what the pieces' denominators must be: one for each factor. That leaves only the numerators unknown, and the equation pins them down, much like matching coefficients anywhere else.

To integrate $\dfrac{P(x)}{Q(x)}$ :

Divide first if $\deg P \ge \deg Q$ , so the remaining fraction is proper. (Adding proper fractions always gives a proper fraction, so an improper one can't be just a sum of pieces.)
Factor $Q(x)$ into linear factors and irreducible quadratics (quadratics with $b^2 - 4ac < 0$ ).
Decompose. Each factor contributes terms:

Factor in $Q(x)$	Terms in the decomposition
$ax + b$	$\dfrac{A}{ax + b}$
$(ax + b)^k$	$\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax + b)^2} + \cdots + \dfrac{A_k}{(ax + b)^k}$
$ax^2 + bx + c$	$\dfrac{Bx + C}{ax^2 + bx + c}$
$(ax^2 + bx + c)^k$	one $\dfrac{B_jx + C_j}{(ax^2 + bx + c)^j}$ for each $j = 1, \dots, k$

Solve for the constants. Multiply through by $Q(x)$ . Plugging in the root of a linear factor wipes out every term except one, since all the others contain that factor (the "cover-up" trick). Matching coefficients of $x^2$ , $x$ and the constant handles whatever's left.
Integrate each piece:

\int \frac{dx}{x - a} = \ln|x - a| + C, \qquad \int \frac{dx}{x^2 + a^2} = \frac1a\arctan\frac{x}{a} + C.

A common mistake. For a repeated factor like $(x - 1)^2$ , it's tempting to write just $\frac{A}{(x - 1)^2}$ . That isn't enough. Combining fractions over $(x - 1)^2$ can produce a numerator with an $x$ in it, and a single constant on top can't match that. You need every power: $\frac{A_1}{x - 1} + \frac{A_2}{(x - 1)^2}$ . In the same way, an irreducible quadratic needs a linear numerator $Bx + C$ , not just a constant.

See it

A population with room for 100 grows by dP/dt = 0.5P(1 − P/100). Separating the variables leaves ∫ dP/(P(100 − P)), which partial fractions split into (1/100)(1/P + 1/(100 − P)), and that gives the logistic curve P = 100/(1 + Ae^(−0.5t)). As you slide the starting population, notice every curve ends up at 100: start below and it rises in an S shape, start above and it sinks back down.

This is (x − 1)/((x − 2)(x − 3)) together with its two pieces, −1/(x − 2) and 2/(x − 3). Notice that each piece is responsible for one of the vertical asymptotes, at x = 2 and x = 3. Add the dashed curves point by point and you get the solid one.

Worked examples

Example 1

Find $\displaystyle\int \frac{x - 1}{x^2 - 5x + 6}\,dx$ .

1
Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$ , so $\dfrac{x - 1}{(x-2)(x-3)} = \dfrac{A}{x - 2} + \dfrac{B}{x - 3}$ .

Example 2

Find $\displaystyle\int \frac{4x^2 - 3x - 4}{x^3 + x^2 - 2x}\,dx$ .

1
The top has lower degree than the bottom, so no division is needed. Factor: $x^3 + x^2 - 2x = x(x + 2)(x - 1)$ , three different linear factors, so write $\dfrac{A}{x} + \dfrac{B}{x + 2} + \dfrac{C}{x - 1}$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\displaystyle\int \frac{dx}{x^2 - 1}$ .

Problem 2+5 XP

Find $\displaystyle\int \frac{3x + 1}{x(x + 1)}\,dx$ .

Problem 3+5 XP

Evaluate $\displaystyle\int_0^3 \frac{dx}{x^2 - 3x - 4}$ .

Problem 4+5 XP

Find $\displaystyle\int \frac{x + 4}{x^3 + x}\,dx$ .

Problem 5+5 XP

Find $\displaystyle\int \frac{dx}{x(x^2 + 1)^2}$ .

Problem 6+5 XP

In a chemical reaction the product's concentration $x$ (mol/L) grows as $\frac{dx}{dt} = 0.5(2 - x)(3 - x)$ , starting from $x = 0$ . Separating gives $t = \int_0^x\frac{du}{0.5(2 - u)(3 - u)}$ . How long until $x = 1$ ?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\displaystyle\int\frac{2x^2 + 3}{(x^2 + 1)^2}\,dx$ .

Where it's used

The logistic population model $P' = rP(1 - P/K)$ is solved by splitting $\frac{1}{P(K - P)}$ into partial fractions. Engineers use the same split to invert Laplace transforms and find how a circuit or suspension responds over time.

Logistic Population GrowthDiff Eq · Biology RLC CircuitsDiff Eq · Electrical Engineering

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