Partial fractions (HL)

Contents

Overview

When dividing a polynomial by another of higher order, partial fractions can be helpful in integration.

p(x)d(x)=q(x)+r(x)d(x)\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}

Note: if the numerator is of the same or higher order than the denominator, first obtain the quotient q(x)q(x) via polynomial division, prior to using partial fractions on r(x)d(x)\frac{r(x)}{d(x)}.

The first step to partial fractions is to factor the denominator, if it is not already factored.

The correct partial fraction decomposition is valid for all values of xx. Feel free to select values of xx that make certain terms go away.

Case 1: all factors are linear and distinct

Example: Find the partial fraction decomposition of −2x+132x2−5x−3\displaystyle \frac{-2x + 13}{2x^2 -5x - 3}.

The denominator factors as (x−3)(2x+1)(x - 3)(2x + 1).

−2x+132x2−5x−3=Ax−3+B2x+1=A(2x+1)+B(x−3)(x−3)(2x+1)−2x+13=A(2x+1)+B(x−3)\begin{align*}\frac{-2x + 13}{2x^2 -5x - 3} &= \frac{A}{x - 3} + \frac{B}{2x + 1} \\ &= \frac{A(2x + 1) + B(x - 3)}{(x-3)(2x + 1)} \\ -2x + 13 &= A(2x + 1) + B(x - 3) \end{align*}

When x=3x = 3,

−2(3)+13=7AA=1\begin{align*}-2(3)+13 &= 7A \\ A &= 1 \end{align*}

When x=−12x = -\frac{1}{2},

14=−7B2−4=B\begin{align*}14 &= \frac{-7B}{2} \\ -4 &= B \end{align*}
−2x+132x2−5x−3=1x−3−42x+1■\frac{-2x + 13}{2x^2 -5x - 3} = \frac{1}{x - 3} - \frac{4}{2x + 1} \qed

Case 2: factors are linear but not distinct

This is not useful (in this course) since the original function can be easily integrated. So it is beyond the syllabus.

Example: Find the partial fraction decomposition of 4x−11(x−2)2\displaystyle \frac{4x - 11}{(x-2)^2}.

4x−11(x−2)2=Ax−2+B(x−2)2=A(x−2)+B(x−2)24x−11=A(x−2)+B\begin{align*} \frac{4x - 11}{(x-2)^2} &= \frac{A}{x - 2} + \frac{B}{(x-2)^2} \\ &= \frac{A(x-2) + B}{(x-2)^2} \\ 4x - 11 &= A(x - 2) + B \end{align*}

When x=2x = 2,

4(2)−11=−3=B\begin{align*}4(2)-11 = -3 &= B \end{align*}

BB is not multiplied by a factor so just use any number, for example x=3x = 3:

1=A+(−3)4=A\begin{align*} 1 &= A + (-3) \\ 4 &= A \end{align*}
4x−11(x−2)2=4x−2−3(x−2)2■\frac{4x - 11}{(x-2)^2}= \frac{4}{x - 2} - \frac{3}{(x-2)^2} \qed

Notes

If integrating, and the denominator cannot be factored using real numbers, then complete the square.

∫1x2+a2 dx=1aarctan⁡xa+C\int \frac{1}{x^2 + a^2}\d x = \frac 1a \arctan \frac xa + C

This technique is discussed in rational functions integration.

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