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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 whole skill.

Year 12-13EDEXCEL 9MA0 8.6

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.

WHAT YOU PROBABLY THINK

∫ 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 the split earns its keep, because each linear piece integrates to a logarithm on sight, with the bottom's own coefficient dividing,

∫ 23x + 5 dx = 23ln|3x + 5| + c

which differentiating back confirms, the chain rule's 3 cancelling the ⅔.

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, which is the standing shape of this topic.

The curve x plus 7 over x minus 1 times x plus 3, with the region from 2 to 5 shaded: partial fractions split the integrand into 2 over x minus 1 minus 1 over x plus 3, and the two logs combine to give an area of exactly the natural log of 10y = (x + 7)/((x − 1)(x + 3))area = ln 1025split into 2/(x − 1) − 1/(x + 3), then two logs
FIG. 1The same integrand as a definite integral: the area under (x + 7)/((x − 1)(x + 3)) from 2 to 5 comes out as ln 10, the two logs folded into one.

YOUR TURN

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 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 ask for a single log, and the log laws from the exponentials unit are what compress four terms into one.

When the split is the wrong tool

Now for the lie at the top. ∫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 claim collapses: 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.

Three rational integrands and their integrals: 2 over 3x plus 5 gives two thirds log of the bottom, x over x squared plus 5 gives half the log of the bottom, and 2 over 2x minus 1 to the fourth is a power, not a log, giving minus a third of the bracket to the minus 32/(3x + 5)⅔ ln|3x + 5| + cadjust for the 3x/(x² + 5)½ ln(x² + 5) + cf’/f, half of it2/(2x − 1)⁴−⅓(2x − 1)⁻³ + cpower, not loglook at the bottom before reaching for a logarithm
FIG. 2Three bottoms, three verdicts: a linear bottom logs with an adjustment, an f'/f pair logs with a half, and a bracket to the fourth is power-rule territory.

YOUR TURN

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. Reading the bottom, and its relationship to the top, chooses the method.

TRY IT UNSEEN

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. Expecting a log from every piece is the same lie in a new spot.

THE EXAM BIT

  • Algebra first: split into partial fractions before any integrating, and quote the split as working; 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.
  • Definite log answers compress by the log laws, and questions usually say 'in the form ln k' to signal it.
  • 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.

CHECK YOUR PROGRESS

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  • 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.

No animated video for this topic yet; these notes stand alone.