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Appendix 10.A Partial Fraction Decomposition

Partial fraction decomposition is, essentially, the inverse operation of combining fractions by putting them over a common denominator. More formally, partial fraction decomposition expresses a proper rational function (i.e. a function that is the ratio of two polynomials where the degree of the polynomial in the denominator is greater that the degree of the polynomial in the numerator) as the sum of proper rational functions of lesser degree.

Write \(\dfrac{2}{x+2}-\dfrac{3}{2x-1}\) as a single fraction.

Answer.

\(=\dfrac{x-8}{(x+2)(2x-1)}\)

Solution.

Using \((x+2)(2x-1)\) as the common denominator we get

\begin{align*} \frac{2}{x+2}-\frac{3}{2x-1} \amp=\frac{2(2x-1)}{(x+2)(2x-1)}-\frac{3(x+2)}{(x+2)(2x-1)}\\ \amp =\frac{4x-2-3x-6}{(x+2)(2x-1)}\\ \amp =\frac{x-8}{(x+2)(2x-1)} \end{align*}

In partial fraction decomposition, for each distinct linear factor \((ax+b)\) in the denominator include a term \(\dfrac{A}{ax+b}\) in the decomposition, where \(A\) is a value we have to determine.

Find the partial fraction decomposition of \(\dfrac{x-8}{(x+2)(2x-1)}\text{.}\)

Answer.
\(\dfrac{x-8}{(x+2)(2x-1)}=\dfrac{2}{x+2}-\dfrac{3}{2x-1}\)
Solution.

Because the denominator of this rational function contains two linear polynomial terms the partial fraction decomposition takes the form

\begin{equation} \frac{x-8}{(x+2)(2x-1)}=\frac{A}{x+2}+\frac{B}{2x-1}\label{Eqn_example11_4}\tag{10.A.4} \end{equation}

To determine the values for \(A\) and \(B\) multiply both sides of (10.A.4) by \((x+2)(2x-1)\text{.}\) This gives

\begin{align*} x-8 \amp=A(2x-1)+B(x+2)\\ \amp =(2A+B)x+(2B-A) \end{align*}
For the polynomials on both sides of the equation to be equal they must have the same coefficients and so
\begin{align*} 2A+B \amp=1\\ -A+2B \amp=-8 \end{align*}
Solving these simultaneous equations gives \(A=2,\hspace{2mm} B=-3\) and hence

\begin{equation*} \frac{x-8}{(x+2)(2x-1)}=\frac{2}{x+2}-\frac{3}{2x-1} \end{equation*}

If the denominator of the rational function contains a linear factor to some power, i.e. \((ax+b)^n\text{,}\) then the partial fraction decomposition should contain the terms

\begin{equation*} \frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+...+\frac{A_n}{(ax+b)^n} \end{equation*}

If the denominator of the rational function contains a quadratic factor, i.e. \((ax^2+bx+c)\text{,}\) then the partial fraction decomposition should contain the term

\begin{equation*} \frac{Ax+B}{ax^2+bx+c} \end{equation*}

Find the partial fraction decomposition of \(\dfrac{2x^3-5x^2+2x-2}{(x^2+2)(x-1)^2}\text{.}\)

Answer.
\(\dfrac{2x^3-5x^2+2x-2}{(x^2+2)(x-1)^2}=\dfrac{2x}{x^2+2}-\dfrac{1}{(x-1)^2}\)
Solution.

Because the denominator contains a quadratic term a repeated linear term the partial fraction decomposition takes the form

\begin{equation} \frac{2x^3-5x^2+2x-2}{(x^2+2)(x-1)^2}=\frac{Ax+B}{x^2+2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}\label{Eqn_example12_5}\tag{10.A.5} \end{equation}

Multiply both sides of (10.A.5) by \((x^2+2)(x-1)^2\) and collect like terms. This gives

\begin{equation*} 2x^3-5x^2+2x-2=(A+C)x^3+(-2A+B-C+D)x^2+(A-2B+2C)x+(B-2C+2D) \end{equation*}

Equating the coefficients of the polynomials on each side gives

\begin{align*} A+C \amp =2\\ -2A+B-C+D \amp=-5\\ A-2B+2C \amp =2\\ B-2C+2D \amp=-2 \end{align*}
Solving these simultaneous equations gives \(A=2\text{,}\) \(B=0\text{,}\) \(C=0\text{,}\) \(D=-1\) and hence

\begin{equation*} \frac{2x^3-5x^2+2x-2}{(x^2+2)(x-1)^2}=\frac{2x}{x^2+2}-\frac{1}{(x-1)^2} \end{equation*}

Exercises Example Tasks

1.

Evaluate the integral

\begin{equation*} \int \frac{x^2-2x+3}{(x^2+1)(x-1)}\hspace{2mm} dx \end{equation*}