Maths › Integration › Integrating rational functions
Integrating rational functions
A fraction of polynomials rarely integrates as it stands, but algebra can always reshape it until it does. Partial fractions split one stubborn fraction into pieces that each make a logarithm, while other bottoms hide a reversed chain or a plain negative power, and telling the cases apart is the skill being tested.
Builds on Partial fractions and Integration by substitution and by parts.
IN THIS TOPIC
- Split a rational integrand into partial fractions and integrate each piece.
- Recognise f'/f numerators and negative-power forms that need no split at all.
- Evaluate definite integrals of rational functions, compressing the answer with log laws.
COMMON MISCONCEPTION
∫ 1/f(x) dx = ln|f(x)| + c.
Split, then integrate
No shelf entry covers a general fraction of polynomials, and the way in is algebra done first. The partial fractions lesson taught the split. This lesson is where that split is put to use, because each linear piece integrates to a logarithm directly, with the denominator's own coefficient dividing.
Differentiate that back and the chain rule's 3 cancels the ⅔, which is the whole reason the adjustment is there.
WORKED EXAMPLE
Two logs from one fraction
Find ∫(x + 7)/((x − 1)(x + 3)) dx.
Split first. (x + 7)/((x − 1)(x + 3)) = A/(x − 1) + B/(x + 3), and the cover-up method gives A = 2 and B = −1.
Each piece is a log: ∫(2/(x − 1) − 1/(x + 3)) dx.
The integral is 2 ln|x − 1| − ln|x + 3| + c.
All the calculus took one line. The algebra before it was the actual work, and it is where the marks are.
GUIDED PRACTICE
The definite version
Using the split above, find the exact value of ∫25 (x + 7)/((x − 1)(x + 3)) dx, before opening the working.
Show the working
The brackets are [2 ln(x − 1) − ln(x + 3)] from 2 to 5. No modulus is needed, since both bottoms stay positive across the interval.
At 5: 2 ln 4 − ln 8 = ln 16 − ln 8 = ln 2. At 2: 2 ln 1 − ln 5 = −ln 5.
Subtracting, ln 2 + ln 5 = ln 10.
Papers want a single log. The log laws from the exponentials unit are what compress four terms into one, and a phrase like “in the form ln k” in the question is telling you to do it.
When the split is the wrong tool
∫x/(x2 + 5) dx really is ½ ln(x2 + 5) + c, but only because the numerator is half the derivative of the bottom, the f'/f pattern from the substitution lesson. Drop the x from the top and the pattern is gone. Differentiating ln(x2 + 5) produces 2x/(x2 + 5), not 1/(x2 + 5). A logarithm appears when the top is the bottom's derivative up to a constant, and at no other time.
GUIDED PRACTICE
Two without a split
Find ∫x/(x2 + 5) dx and ∫2/(2x − 1)4 dx.
Show the working
The first is f'/f with a factor missing: ½ ln(x2 + 5) + c, no modulus required because x2 + 5 is never negative.
The second rewrites as 2(2x − 1)−4, a power, and reversing the power rule with the inner 2 dividing gives −1/(3(2x − 1)3) + c.
Neither needed partial fractions. Read the bottom, and read its relationship to the top, and the method chooses itself.
INDEPENDENT PRACTICE
A repeated factor
Find ∫(3x + 1)/(x + 1)2 dx.
Show the working
The repeated-factor template applies: (3x + 1)/(x + 1)2 = A/(x + 1) + B/(x + 1)2, and matching gives A = 3, B = −2.
The first piece logs. The second is a negative power, integrating to +2/(x + 1).
The integral is 3 ln|x + 1| + 2/(x + 1) + c.
One split produced one log and one power. A log appears only where the top is the bottom's derivative, and not every piece obliges.
ASSESSMENT FOCUS
- Algebra first. Split into partial fractions before any integrating, and quote the split as working, because it carries its own marks.
- A linear bottom gives a log divided by the bottom's coefficient. Differentiate back to check the adjustment.
- A log arises only when the top is the bottom's derivative up to a constant. Otherwise suspect a negative power.
- Keep the modulus in ln|x − a| unless the interval keeps the bracket positive, and say so when you drop it.
CHECK YOURSELF
Find the exact value of ∫02 2/((x + 1)(x + 3)) dx, giving the answer as a single logarithm.
Show a hint
Cover-up gives 1/(x + 1) − 1/(x + 3); then log laws.
Show the answer
The split is 2/((x + 1)(x + 3)) = 1/(x + 1) − 1/(x + 3).
The brackets give [ln(x + 1) − ln(x + 3)] from 0 to 2 = (ln 3 − ln 5) − (0 − ln 3).
That is 2 ln 3 − ln 5 = ln(9/5).
Split, two logs, compress. The full method of the lesson in three lines.
Split a rational integrand by partial fractions; every linear piece becomes a logarithm.
Logs come from f'/f and nowhere else; other bottoms are powers written in fraction notation.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the integrating rational functions questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Split a rational integrand into partial fractions and integrate each piece.
- Recognise f'/f numerators and negative-power forms that need no split at all.
- Evaluate definite integrals of rational functions, compressing the answer with log laws.
Open the full revision checklist to see every objective in the course in one place.