4.08f - Integrating with partial fractions | OCR A-Level Further Mathematics A

Use partial fractions to integrate rational functions with linear factors, repeated linear factors and irreducible quadratic factors. You will combine logarithms with inverse tangent terms, divide improper fractions before integrating, and check the whole antiderivative by differentiation.

Integrating linear partial fractions

Why does $\displaystyle\int\frac{x^3+2x^2+3x+4}{x^2+1}\,dx$ need more than a logarithm? Its numerator is too large for the fraction to be proper, and its denominator is not the derivative of its numerator. Partial fractions and division will reveal three different kinds of term in one answer.

You already know

Choose the partial-fraction form from the denominator, clear denominators and solve the resulting identity. Include every power of a repeated linear factor, use a linear numerator over an irreducible quadratic, and divide first if the numerator's degree is at least the denominator's. These decomposition rules are not in the formula booklet.

The new task is to integrate each term of the decomposition. For a simple linear denominator, reverse logarithmic differentiation. In particular, $\int 6/(2x+1)\,dx=3\ln|2x+1|+c$ , because differentiating the logarithm introduces a factor of $2$ .

Integrating a simple linear fraction

For real constants $a\ne0$ , $b$ and $k$ , on an interval where $ax+b\ne0$ ,

\int\frac{k}{ax+b}\,dx=\frac{k}{a}\ln|ax+b|+c

The general rule $\int f'(x)/f(x)\,dx=\ln|f(x)|+c$ is printed; this is its linear case.
In the formula booklet.

The modulus lets the logarithmic expression work on either side of a zero of the linear factor. It does not make an antiderivative valid across that zero. Constants may be chosen separately on disconnected intervals.

Worked example 1: two simple factors

Question. Find $\displaystyle\int\frac{3x+7}{(x+1)(x+3)}\,dx$ .

Plan. Find the two constant numerators, then integrate the two simple fractions.

\begin{aligned} 3x+7&=A(x+3)+B(x+1) && \text{clear denominators} \\ x=-1&\Rightarrow 4=2A\Rightarrow A=2 && \text{remove the }B\text{ term} \\ x=-3&\Rightarrow -2=-2B\Rightarrow B=1 && \text{remove the }A\text{ term} \\ I&=\int\left(\frac2{x+1}+\frac1{x+3}\right)dx && \text{decompose} \\ &=2\ln|x+1|+\ln|x+3|+c && \text{integrate term by term} \end{aligned}

Why this step? Each logarithm differentiates to its own simple fraction, so splitting the denominator makes the integral immediately recognisable.

Answer. $2\ln|x+1|+\ln|x+3|+c$ , on any interval avoiding $-3$ and $-1$ .

Check. Recombining its derivative gives $[2(x+3)+(x+1)]/((x+1)(x+3))$ , exactly the original integrand.

Change only the numerator from $3x+7$ to $5x+9$ . The factor structure and integration methods stay fixed; only the constants need to change.

Worked example 2: a changed numerator

Question. Find $\displaystyle\int\frac{5x+9}{(x+1)(x+3)}\,dx$ .

Plan. Use the same two-fraction form and determine its new constants.

\begin{aligned} 5x+9&=A(x+3)+B(x+1) && \text{cleared identity} \\ x=-1&\Rightarrow A=2 && 4=2A \\ x=-3&\Rightarrow B=3 && -6=-2B \\ I&=\int\left(\frac2{x+1}+\frac3{x+3}\right)dx && \text{substitute the constants} \\ &=2\ln|x+1|+3\ln|x+3|+c && \text{logarithmic integrals} \end{aligned}

Why this step? The numerator affects the weights of the two logarithms, not which logarithms appear.

Answer. $2\ln|x+1|+3\ln|x+3|+c$ , on intervals avoiding the two poles.

Check. $2(x+3)+3(x+1)=5x+9$ , so differentiating and recombining recovers the numerator.

In a definite integral, evaluate only when the whole integration interval avoids a pole, or handle it as an improper integral. For example, the first antiderivative is valid throughout $[0,1]$ , and upper minus lower gives $\ln(16/3)$ . Logarithm laws used to combine exact endpoint terms are assumed knowledge and are not printed in the booklet.

Watch out

Wrong: $\int dx/((x+1)(x+3))=\ln|(x+1)(x+3)|+c$ . The logarithmic derivative would have numerator $2x+4$ , not $1$ . Decompose first unless the numerator is already the denominator's derivative times a constant.

Keep the factors and change the numerator once more.

Repeated factors give powers as well as logarithms

A repeated linear factor contributes several powers to a decomposition. Only its first power gives a logarithm; the higher powers follow the ordinary power rule.

For example, $(2x+1)^{-2}$ integrates to $-\frac12(2x+1)^{-1}$ , because differentiating the proposed answer multiplies by $-1$ and then by $2$ . The exceptional power is $-1$ , for which the ordinary power-rule denominator would be zero.

Integrating higher powers of a linear factor

For $a\ne0$ and an integer $r\ge2$ , where $ax+b\ne0$ ,

\int\frac{k}{(ax+b)^r}\,dx =\frac{k}{a(1-r)}(ax+b)^{1-r}+c

This follows from the printed rule for integrating $f'(x)(f(x))^n$ ; $r=1$ instead needs a logarithm.
In the formula booklet.

Keep the two causes of a coefficient separate: the exponent changes by one, and the inner linear scale must be divided out.

Worked example 3: a repeated factor

Question. Find $\displaystyle\int\frac{6x+7}{(2x+1)^2}\,dx$ .

Plan. Write the fraction using both powers of $2x+1$ , then use a logarithm and a power integral.

\begin{aligned} 6x+7&=A(2x+1)+B && \text{clear the fractions} \\ A&=3,\quad B=4 && \text{compare coefficients} \\ I&=\int\left(\frac3{2x+1}+\frac4{(2x+1)^2}\right)dx && \text{both powers present} \\ &=\frac32\ln|2x+1|-\frac2{2x+1}+c && \text{integrate} \end{aligned}

Why this step? The squared denominator is not a logarithmic term. Its primitive has power $-1$ , with a negative coefficient.

Answer. $\frac32\ln|2x+1|-2/(2x+1)+c$ , on intervals avoiding $-1/2$ .

Check. Its derivative is $3/(2x+1)+4/(2x+1)^2$ , and the combined numerator is $6x+7$ .

Change only the constant in the numerator from $7$ to $11$ . The logarithmic coefficient stays fixed, while the repeated-factor term doubles.

Worked example 4: changing the repeated-factor contribution

Question. Find $\displaystyle\int\frac{6x+11}{(2x+1)^2}\,dx$ .

Plan. Find the same two constants again, then change the rational term in the primitive.

\begin{aligned} 6x+11&=A(2x+1)+B && \text{cleared identity} \\ A&=3,\quad B=8 && \text{compare coefficients} \\ I&=\int\left(\frac3{2x+1}+8(2x+1)^{-2}\right)dx && \text{rewrite the higher power} \\ &=\frac32\ln|2x+1|-\frac4{2x+1}+c && \text{divide by the inner scale} \end{aligned}

Why this step? Integrating the second term divides $8$ by $2\times(-1)$ , so its coefficient is $-4$ , not $-8$ .

Answer. $\frac32\ln|2x+1|-4/(2x+1)+c$ .

Check. Differentiating the rational term gives $8/(2x+1)^2$ ; combining with the logarithmic derivative gives $6x+11$ above the original denominator.

The same principle applies to a higher repeated power when it occurs: retain all terms in the decomposition, then integrate each power separately.

Watch out

Wrong: $\int(2x+1)^{-2}\,dx=\ln|(2x+1)^2|+c$ . A squared denominator uses the power rule, giving $-1/(2(2x+1))+c$ . Only exponent $-1$ gives a logarithm.

Change the numerator constant again while preserving the repeated factor.

Splitting the numerator over a quadratic

Above an irreducible quadratic, the linear numerator has two jobs: its variable term matches a logarithmic derivative, while its constant term gives inverse tangent.

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4.08f - Integrating with partial fractions

Integrating linear partial fractions

Repeated factors give powers as well as logarithms

Splitting the numerator over a quadratic

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