Integration Using Partial Fractions (College Board AP® Calculus BC): Revision Note

Integration using partial fractions

What are partial fractions?

• A rational function $\frac{f\left(x\right)}{g\left(x\right)}$ can be written as the sum of partial fractions provided:

  • $f\left(x\right)$ and $g\left(x\right)$ are polynomials

  • the degree of $f$ is less than the degree of $g$

• Each partial fraction has the following properties:

  • the denominator is a factor of $g\left(x\right)$

  • the degree of the denominator will be less than the degree of $g$

  • the degree of the numerator will be less than the degree of the denominator

• In this course $g\left(x\right)$ will be a product of distinct linear factors

  • Usually only two factors

  • The numerators of the partial fractions will be constant

• For example, $\frac{8x+10}{\left(x+3\right)\left(2x-1\right)}=\frac{2}{x+3}+\frac{4}{2x-1}$

How can I write a rational function as a sum of partial fractions?

• STEP 1
  Factor the denominator

  • e.g. $\frac{2x-17}{x^2-3x-10}=\frac{2x-17}{\left(x+2\right)\left(x-5\right)}$

• STEP 2
  Write as a sum of partial fractions

  • The numerators are unknown constants

  • The denominators are the linear factors

    • e.g. $\frac{2x-17}{\left(x+2\right)\left(x-5\right)}=\frac{A}{x+2}+\frac{B}{x-5}$

• STEP 3
  Multiply both sides by the denominator of the original fraction

  • This gets rid of all the fractions

    • e.g. $\frac{\left(2x-17\right)\cancel{\left(x+2\right)}\cancel{\left(x-5\right)}}{\cancel{\left(x+2\right)}\cancel{\left(x-5\right)}}=\frac{A\cancel{\left(x+2\right)}\left(x-5\right)}{\cancel{x+2}}+\frac{B\left(x+2\right)\cancel{\left(x-5\right)}}{\cancel{x-5}}$

    • which simplifies to $2x-17=A\left(x-5\right)+B\left(x+2\right)$

• STEP 4
  Find the values of the unknown constants

  • One method is to substitute the roots of the denominators into the equation

    • e.g. Substitute $x=5$

      $\begin{array}{rcl}x& =& 5:\\ 2\left(5\right)-17& =& A\left(5-5\right)+B\left(5+2\right)\\ -7& =& 7B\\ -1& =& B\end{array}$

    • e.g. Substitute $x=-2$

      $\begin{array}{rcl}x& =& -2:\\ 2\left(-2\right)-17& =& A\left(-2-5\right)+B\left(-2+2\right)\\ -21& =& -7A\\ 3& =& A\end{array}$

  • An alternative method is to compare the coefficients of the equation

    • e.g. Collect like-terms on the right-hand side
      $2x-17=\left(A+B\right)x+\left(-5A+2B\right)$

    • Form two simultaneous equations

      $\begin{array}{rcl}A+B& =& 2\\ -5A+2B& =& -17\end{array}$

    • Solve to get $A=3$ and $B=-1$

• STEP 5
  Write out the partial fractions

  • e.g. $\frac{2x-17}{\left(x+2\right)\left(x-5\right)}=\frac{3}{x+2}-\frac{1}{x-5}$

Can I use partial fractions if the degree of the numerator is not smaller than the degree of the denominator?

• You can use long division to write a rational function as the sum of a polynomial and another rational function

  • e.g. $\frac{2x^3-x^2-x-7}{x^2-x-6}=2x+1+\frac{12x-1}{x^2-x-6}$

• You can then write the new rational function as a sum of partial fractions

  • e.g. $\frac{2x^3-x^2-x-7}{x^2-x-6}=2x+1+\frac{7}{x-3}+\frac{5}{x+2}$

How do I integrate using partial fractions?

• It is straightforward to integrate a rational function if it is written as the sum of partial fractions

• Integrate each partial fraction separately

  • $\int \frac{1}{ax+b}dx=\frac{1}{a}\ln \left|ax+b\right|+C$