A rational function such as (3x + 1)/((x − 1)(x + 1)) cannot be integrated as it stands, but once it is split into partial fractions each term integrates to a logarithm: ∫ A/(x − a) dx = A ln|x − a| + c. For definite integrals, use the laws of logarithms to combine the answer into a single exact log.
Key results
- ∫ A/(x − a) dx = A ln|x − a| + c
- ∫ 1/(ax + b) dx = (1/a)ln|ax + b| + c
- ln p − ln q = ln(p/q) and ln p + ln q = ln(pq)
- k ln p = ln(pᵏ)
If the fraction is improper, divide first; the quotient integrates to a polynomial.
Integrating a sum of partial fractions
Once a fraction is written as a sum of terms like A/(x − a), each term integrates to a logarithm: A ln|x − a|. A constant numerator simply multiplies the log. If the denominator is ax + b with a ≠ 1, divide by a. Keep the modulus signs in indefinite integrals.
Finding the partial fractions first
Write the fraction in the form A/(x − a) + B/(x − b), multiply through by the denominator, and substitute x = a and x = b to find each constant in turn. Substituting the value that makes one factor zero isolates the other constant. Then integrate term by term.
Definite integrals and exact logs
Evaluate between the limits, then combine the logarithms using ln p − ln q = ln(p/q), ln p + ln q = ln(pq) and k ln p = ln(pᵏ). The final answer is usually a single logarithm of a fraction, such as ln(9/8). You can combine the logs before or after substituting the limits.
Harder denominators
You may need to factorise the denominator first, deal with three linear factors, or handle a repeated factor, where a term like B/(x − a)² integrates to −B/(x − a) rather than to a log. If the fraction is improper, carry out the division first and integrate the polynomial part separately.
Areas under rational curves
The exact area under a rational curve between two limits is found in the same way: split, integrate, evaluate, combine. Make sure the curve is positive across the interval, so the integral gives the area rather than a signed value.
Worked examples
Example 1. Find ∫ (2/(x − 3) + 1/(x + 1)) dx.
- 2/(x − 3) integrates to 2 ln|x − 3|.
- 1/(x + 1) integrates to ln|x + 1|.
Answer: 2 ln|x − 3| + ln|x + 1| + c.
Marks: A1 for both terms and + c.
Example 2. Find ∫ (3x + 1)/((x − 1)(x + 1)) dx.
- Write (3x + 1)/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1), so 3x + 1 = A(x + 1) + B(x − 1).
- x = 1: 4 = 2A, so A = 2. x = −1: −2 = −2B, so B = 1.
- Integrate 2/(x − 1) + 1/(x + 1).
Answer: 2 ln|x − 1| + ln|x + 1| + c.
Marks: M1 for finding A = 2 and B = 1, A1 for the integral.
Example 3. Find the exact value of ∫ 1/((x + 1)(x + 2)) dx from 1 to 2.
- Split: 1/((x + 1)(x + 2)) = 1/(x + 1) − 1/(x + 2).
- Integrate: [ln(x + 1) − ln(x + 2)] = [ln((x + 1)/(x + 2))] from 1 to 2.
- ln(3/4) − ln(2/3) = ln(3/4 × 3/2).
Answer: ln(9/8).
Marks: M1 for splitting and integrating, A1 for ln(9/8).
Example 4. Find ∫ (1/x + 2/(x − 1) + 1/(x + 1)) dx.
- Three factors give three logarithms.
- ln|x| + 2 ln|x − 1| + ln|x + 1|.
Answer: ln|x| + 2 ln|x − 1| + ln|x + 1| + c.
Marks: A1 for all three terms.
Example 5. Find the exact area under y = 1/((x + 1)(x + 2)) from x = 0 to x = 1.
- Split: 1/(x + 1) − 1/(x + 2).
- Integrate: [ln((x + 1)/(x + 2))] from 0 to 1.
- ln(2/3) − ln(1/2) = ln(2/3 × 2).
Answer: ln(4/3) square units.
Marks: M1 for splitting and integrating, A1 for ln(4/3).
Common mistakes
- Integrating before splitting. Split into partial fractions first; the combined fraction has no standard integral.
- Losing the constant numerator. 2/(x − 1) integrates to 2 ln|x − 1|, not ln|x − 1|.
- Forgetting to divide by a. 1/(2x + 1) integrates to ½ ln|2x + 1|.
- Combining logs incorrectly. ln p − ln q = ln(p/q), not ln p / ln q.
- Skipping the division for an improper fraction. If the top degree is not lower than the bottom, divide first.
How integration with partial fractions is examined
This is Pure 4 content (WMA14) in Edexcel International A-Level, and Year 2 Pure in Edexcel A-Level (9MA0). A question usually has two parts:
- express a fraction in partial fractions
- then integrate it, often as a definite integral
- give the answer as a single logarithm or in the form ln k
- use partial fractions to solve a differential equation
Two-part questions of this kind, with mark schemes, appear in the Pure 4 integration exam questions. The Pure 4 integration worksheet and the integration revision engine give extra practice.
The splitting step is covered in partial fractions. The integrals rely on integrating 1/(ax + b) and the laws of logarithms.
Frequently asked questions
Why do partial fractions help with integration?
They turn one fraction with no standard integral into a sum of simple fractions, each of which integrates to a logarithm.
What is the integral of 1/(x − a)?
ln|x − a| + c.
How do I give an exact answer to a definite integral with logs?
Use the laws of logarithms to combine the terms into a single log, such as ln(9/8).