Integrating with Partial Fractions (DP IB Analysis & Approaches (AA): HL): Revision Note

Integrating with partial fractions

What are partial fractions?

• Partial fractions arise when a quotient is rewritten as the sum of fractions

  • The process is the opposite of adding or subtracting fractions

• Each partial fraction has a denominator which is a linear factor of the quotient’s denominator

  • e.g.  A quotient with a denominator of $x^2+4x+3$

    • factorises to $\left(x+1\right)\left(x+3\right)$

    • so the quotient will split into two partial fractions

    • one with the (linear) denominator $\left(x+1\right)$

    • one with the (linear) denominator $\left(x+3\right)$

How do I know when to use partial fractions in integration?

• For this course, the denominators of the quotient will be of quadratic form

  • i.e. $f\left(x\right)=a{x}^2+bx+c$

• However check to see if the quotient can be written in the form $\frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}$

  • In this case, reverse chain rule applies

• If the denominator does not factorise then the inverse trigonometric functions may be involved

How do I integrate using partial fractions?       

• STEP 1
  Rewrite the quotient in the integrand as the sum of partial fractions
  This involves factorising the denominator, writing it as an identity of two partial fractions and solving to find their numerators

  • e.g. $I=\int \frac{1}{x^2+4x+3}dx=\int \frac{1}{\left(x+1\right)\left(x+3\right)}dx=\frac{1}{2}\int \left(\frac{1}{x+1}-\frac{1}{x+3}\right)dx$

• STEP 2
  Integrate each partial fraction, leading to an expression involving the sum or difference of natural logarithms

  • e.g. $I=\frac{1}{2}\int \left(\frac{1}{x+1}-\frac{1}{x+3}\right)dx=\frac{1}{2}\left(\ln \left|x+1\right|-\ln \left|x+3\right|\right)+c$

• STEP 3
  Use the laws of logarithms to simplify the expression and/or apply the integration limits
  (Simplifying first may make applying the limits easier)

  • e.g. $I=\frac{1}{2}\ln \left|\frac{x+1}{x+3}\right|+c$

• By rewriting the constant of integration as a logarithm ($c=\ln k$, say) it is also possible to write the final answer as a single term

  • e.g. $I=\frac{1}{2}\ln \left|\frac{x+1}{x+3}\right|+\ln k=\ln \sqrt{\left|\frac{x+1}{x+3}\right|}+\ln k=\ln \left(k\sqrt{\left|\frac{x+1}{x+3}\right|}\right)$