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

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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 squared plus 4 x plus 3

      • factorises to open parentheses x plus 1 close parentheses open parentheses x plus 3 close parentheses

      • so the quotient will split into two partial fractions

      • one with the (linear) denominator open parentheses x plus 1 close parentheses

      • one with the (linear) denominator open parentheses x plus 3 close parentheses

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 parenthesis x right parenthesis equals a x squared plus b x plus c

    • check to see if the quotient can be written in the form fraction numerator f apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction

      • in this case, reverse chain rule applies

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

How do I integrate using partial fractions?       

STEP 1
Write 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 using values of x to find their numerators
e.g. I equals integral subscript blank superscript blank fraction numerator 1 over denominator x squared plus 4 x plus 3 end fraction space d x equals integral subscript blank superscript blank fraction numerator 1 over denominator open parentheses x plus 1 close parentheses open parentheses x plus 3 close parentheses end fraction space d x equals 1 half integral subscript blank superscript blank open parentheses fraction numerator 1 over denominator x plus 1 end fraction minus fraction numerator 1 over denominator x plus 3 end fraction close parentheses space d x

STEP 2
Integrate each partial fraction leading to an expression involving the sum of natural logarithms
e.g. I equals 1 half integral subscript blank superscript blank open parentheses fraction numerator 1 over denominator x plus 1 end fraction minus fraction numerator 1 over denominator x plus 3 end fraction close parentheses space d x equals 1 half open square brackets ln space open vertical bar x plus 1 close vertical bar minus ln space open vertical bar x plus 3 close vertical bar close square brackets plus c

STEP 3
Use the laws of logarithms to simplify the expression and/or apply the limits
(Simplifying first may make applying the limits easier)
e.g. I equals 1 half ln space open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar plus c

  • By rewriting the constant of integration as a logarithm (c equals ln space k, say) it is possible to write the final answer as a single term
    e.g. I equals 1 half ln space open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar plus ln space k equals ln space square root of open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar end root plus ln space k equals ln space open parentheses k square root of open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar end root close parentheses

Examiner Tips and Tricks

  • Always check to see if the numerator can be written as the derivative of the denominator

    • If so then it is reverse chain rule, not partial fractions

    • Use the number of marks a question is worth to help judge how much work should be involved

Worked Example

Find integral subscript blank superscript blank fraction numerator 3 x plus 1 over denominator x squared plus 3 x minus 10 end fraction space d x.

5-9-3-ib-hl-aa-only-we-soltn
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Paul

Author: Paul

Expertise: Maths Content Creator (Previous)

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams.

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