Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

2.5: Partial Fractions

The Fundamental Theorem of Algebra

The fundamental theorem of algebra states that if \(P(x)\) is a polynomial of degree \(n\) then \(P(x)\) can be factored into linear factors over the complex numbers. Furthermore, \(P(x)\) can be factored over the real numbers as a product of linear and quadratic terms and any rational function can be split up as a sum of rational function with denominators of the form

\[ (x - r)n  \]

or

\[  x^2 + Ax + B. \]

Example 1: Partial Fractions

Consider the rational function

\[ P(x) =\dfrac{3x + 2 }{x^2 - 1} = \dfrac{3x +2}{(x - 1)(x + 1)}. \]

We want to write it in the form

\[ \dfrac{ 3x + 2}{(x - 1)(x + 1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}. \]

To do this we need to solve for \(A\) and \(B\). Multiplying by the common denominator

\[(x-1)(x+1) \]

we have

\[  3x  + 2 = A(x + 1) + B(x - 1).  \]

Now let \(x = 1\)

\(5 = 2A + 0\)

\(A = \dfrac{5}{2}\).

Now let \(x = -1\)

\(-1 = -2B\)

\(B = \dfrac{1}{2}\).

Hence we can write

\[ \dfrac{ 3x + 2}{ x^2 - 1} = \dfrac{\frac{5}{2}}{x-1} + \dfrac{\frac{1}{2}}{x+1}.\]

This is called the partial fraction decomposition of \(P(x)\).

Example 2

Find the Partial Fraction Decomposition of

\[ P(x) = \dfrac{3x^2 + 4x + 7}{x^3 - 2x^2 + x} = \dfrac{3x^2 + 4x + 7}{x(x-1)^2}. \]

We write

\[ \dfrac{3x^2 + 4x + 7}{x(x-1)^2} = \dfrac{A}{x - 1 } + \dfrac{B}{(x - 1)^2} + \dfrac{C}{x}. \]

Multiplying by the common denominator, we have

\[ A(x)(x - 1) + Bx + C(x - 1)2  = 3x^2 + 4x + 7. \]

Let \(x = 0\):

\(C = 7\).

Let \(x = 1\):

We have

\(B = 14\).

Now look at the highest degree coefficient:

\[ Ax^2 + Cx^2  = 3x^2.\]

Dividing by \(x^2\) and substituting \(C = 7\):

        \(A + 7 = 3\),     \(A = - 4\).

We conclude that

\[ \dfrac{3x^2 + 4x + 7}{x(x-1)^2} = \dfrac{-4}{x - 1 } + \dfrac{14}{(x - 1)^2} + \dfrac{7}{x}. \]

Example 3

Evaluate

\[  \int \dfrac{x-2}{x(x^2+1)} dx. \]

Solution

We write

\[ \dfrac{x^2 - 2}{x(x^2 +1)} = \dfrac{A}{x} + \dfrac{Bx + C}{x^2 + 1}. \]

Multiplying by the common denominator, we have

\[ A(x^2 + 1) + (Bx + C)x = x^2 - 2.\]

Let \(x = 0\)

\(A = -2\).

Hence

\[ (Bx + C)x = x^2 - 2 + 2x^2 + 2 = 3x^2.\]

So that

\[ Bx^2 + Cx = 3x^2.\]

We see that \(B = 3\) and \(C = 0\)

Hence

\[\begin{align} \int \dfrac{x^2-2}{x(x^2+1)} dx &= \int \Big( \dfrac{-2}{x}+\dfrac{3x}{x^2+1} \Big) dx \\ &= -2\int \dfrac{1}{x} dx +\int \dfrac{3x}{x^2+1} dx \\ &u=x^2+1, \;\;\; du=2x\; dx \\ &= -2\ln|x|+\dfrac{3}{2}\int \dfrac{du}{u} dx \\ &= -2 \ln|x|+\dfrac{3}{2}\ln(x^2+1)+C. \end{align}\]

Example 4

It has been said that the total population of South Lake Tahoe should never exceed 30,000 people. If in 1980 the population was 15,000 and in 2000 it was 20,000, when will the population reach 25,000?

Solution: 

We make the assumption that the rate of increase of the population is proportional to the product of the current population and 30,000 minus the current population. That is:

\[ \dfrac{dP}{dt} = kP (30,000-P).\]

This is a separable differential equation.  Separating gives

\[ \dfrac{dP}{P (30,000-P) } = kdt. \]

Now integrate both sides.  The right hand side is

\[ kt + C.\]

To integrate the right hand side, use partial fractions:

\[ \dfrac{1}{P(30,000 - P)} = \dfrac{A}{P} + \dfrac{B}{30,000 - P}. \]

Multiplying by the common denominator:

        \(1 = A(30,000 - P) + BP \)

        \(P = 0\):      \(1 = 30,000A\)     or     \(A = \dfrac{1}{30,000}\)

        \(P = 30,000\):  \(1 = 30,000B\)     or     \(B = \dfrac{1}{30,000}\).

Now integrate to get

\[ \dfrac{1}{30,000} \ln P - \dfrac{1}{30,000} \ln (30,000 -P) = kt + C \]

or

\[ \ln \dfrac{P}{30,000 -P} = at + b. \]

When \(t = 0\), \(P = 15,000\), so

\[b = 0.\]

When \(t = 20\), \(P = 20,000\), so

\[20a = \ln2.\]

or 

\[ a = \dfrac{\ln \; 2}{20} \]

hence

\[ \ln \dfrac{P}{30,000 -P} = \dfrac{t\; \ln \; 2}{20}. \]

Finally

\[ \ln\; 5 = \dfrac{t \; \ln \; 2}{20} \]

or

\[ t = \dfrac{20 \ln \; 5}{\ln \; 2} = 46.4. \]

The population will reach 2500 in the year 2026.

Problems

Exercises

Find:

1. \[ \int \dfrac{x-2}{x^2+3x-4} dx   \]

2. \[ \int \dfrac{x}{(x+1)(x^2+4)} dx\]

3. \[ \int \dfrac{2x+3}{x^2(x-3)} dx \]

4. \[ \int \dfrac{6x+1}{3x^2+x-2} dx  \]

Contributors

  • Integrated by Justin Marshall.