Mc30 - Integrating rational functions with partial fractions | OCR A-Level Mathematics B (MEI)

Use algebraic decomposition to turn rational integrals into logarithms and powers. Keep repeated factors complete, handle non-unit linear scales and evaluate exact integrals only on valid domain intervals.

Turning distinct linear factors into logarithms

How can a fraction with three denominator factors need only two logarithms? A vanishing partial-fraction coefficient can reveal cancellation. The cancelled denominator’s zero remains excluded from the function as stated. A simpler expression does not automatically have the same domain.

You already know

Clear denominators to obtain a polynomial identity, then find its coefficients. Divide first if the rational expression is improper. Root substitutions in the identity do not make the rational expression defined at its poles.

Once decomposed, each first-power linear denominator gives a logarithm with an inner-derivative scale factor.

A linear denominator

For $a\ne0$ , on an interval avoiding $ax+b=0$ ,

\int\frac{K}{ax+b}\,\mathrm dx=\frac Ka\ln|ax+b|+C.

Derived from the sheet's logarithmic-derivative result; this linear form is not separately printed. Not in the formula sheet: learn it.

For example, $\int3/(1-2x)\,\mathrm dx=-\frac32\ln|1-2x|+C$ . Check a negative inner derivative as carefully as a positive one.

Worked example 1: finding both coefficients

Question. Find $\int(5x+7)/[(x+1)(x+2)]\,\mathrm dx$ .

Plan. Decompose using one constant numerator for each linear factor.

\begin{aligned} 5x+7&=A(x+2)+B(x+1) &&\text{clear denominators}\\ x=-1&\Rightarrow A=2,\quad x=-2\Rightarrow B=3\\ I&=2\ln|x+1|+3\ln|x+2|+C. \end{aligned}

Why substitute the roots? Each makes one term vanish, isolating the other coefficient.

Answer. $2\ln|x+1|+3\ln|x+2|+C$ , separately on intervals excluding $-2,-1$ .

Check. Differentiation recombines to numerator $2(x+2)+3(x+1)=5x+7$ .

Change only the numerator's constant from $7$ to $8$ . The same factors now have different weights.

Worked example 2: a changed constant

Question. Find $\int(5x+8)/[(x+1)(x+2)]\,\mathrm dx$ .

Plan. Recalculate both constants before integrating.

\begin{aligned} 5x+8&=A(x+2)+B(x+1)\\ A&=3,\quad B=2 &&\text{use }x=-1,-2\\ I&=3\ln|x+1|+2\ln|x+2|+C. \end{aligned}

Why no single logarithm of the denominator? That would require a numerator proportional to its derivative, imposing different coefficients.

Answer. $3\ln|x+1|+2\ln|x+2|+C$ .

Check. Its derivative has numerator $3(x+2)+2(x+1)=5x+8$ .

The decomposition $(3x+5)/[(x+1)(x+2)]=2/(x+1)+1/(x+2)$ follows from the same identity method and integrates to two corresponding logarithms.

Watch out

Wrong: remove modulus because the rational function is real. Right: keep modulus unless a specified interval makes the logarithm arguments positive, or use an equivalent positive argument.

Change the numerator's constant again.

Including every power of a repeated factor

A squared linear factor needs both denominator powers in the decomposition. Its second-power fraction integrates to a reciprocal, while its first-power fraction gives a logarithm.

A squared linear denominator

For $a\ne0$ , on an interval avoiding its zero,

\int\frac{K}{(ax+b)^2}\,\mathrm dx=-\frac{K}{a(ax+b)}+C.

Derived from the scaled power rule. Not in the formula sheet: learn it.

First find the partial-fraction coefficient; then apply the separate integration scale factor.

Worked example 3: both denominator powers

Question. Find $\int(3x+2)/[x(x+2)^2]\,\mathrm dx$ .

Plan. Use $A/x+B/(x+2)+D/(x+2)^2$ .

\begin{aligned} 3x+2&=A(x+2)^2+Bx(x+2)+Dx\\ x=0&\Rightarrow A=\frac12,\quad x=-2\Rightarrow D=2\\ A+B&=0\Rightarrow B=-\frac12 &&\text{quadratic coefficient}\\ I&=\frac12\ln|x|-\frac12\ln|x+2|-\frac2{x+2}+C. \end{aligned}

Why include $B/(x+2)$ ? Without it, the cleared numerator's quadratic coefficient cannot cancel.

Answer. The displayed primitive on intervals excluding $0,-2$ .

Check. Recombination gives numerator $(x+2)^2/2-x(x+2)/2+2x=3x+2$ .

Change only the coefficient of $x$ in the repeated factor, from $x+2$ to $2x+2$ . Its root and integration scale both respond to that one change.

Worked example 4: a non-unit repeated factor

Question. Find $\int(3x+2)/[x(2x+2)^2]\,\mathrm dx$ .

Plan. Use the full decomposition and then divide each repeated-factor coefficient by its inner derivative $2$ .

\begin{aligned} 3x+2&=A(2x+2)^2+Bx(2x+2)+Dx\\ A&=\frac12,\quad D=1,\quad B=-1 &&\text{roots and quadratic term}\\ I&=\frac12\ln|x|-\frac12\ln|2x+2|-\frac1{2(2x+2)}+C. \end{aligned}

Why halve the reciprocal coefficient too? The squared-factor power rule also includes the inner derivative.

Answer. The displayed primitive, excluding $0,-1$ .

Check. Differentiating the reciprocal term gives $1/(2x+2)^2$ , the required third fraction.

An example with denominator $(x+1)(x-1)^2$ likewise has coefficients $-1/4,1/4,3/2$ for numerator $2x+1$ , giving $-\frac14\ln|x+1|+\frac14\ln|x-1|-3/[2(x-1)]+C$ . A grouped linear numerator over the squared factor can be equivalent, but must be rewritten before integration.

Exam tip

Write every required denominator power in the setup. MEI schemes can award method credit before a later coefficient slip, but an omitted fraction can prevent a complete valid decomposition.

Change the repeated factor's coefficient again and use a fresh numerator.

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Mc30 - Integrating rational functions with partial fractions

Turning distinct linear factors into logarithms

Including every power of a repeated factor

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