H6 - Integration using partial fractions | AQA A-Level Mathematics

Split rational expressions into fractions you can integrate using logarithms and powers. You will handle distinct and repeated linear factors, prepare an improper fraction and keep the original domain when evaluating definite integrals.

Integrating individual linear-denominator terms

Can $\int\frac{5x+1}{(x-1)(x+2)}\,dx$ be found by taking the logarithm of its denominator? Differentiating that denominator gives $2x+1$ , which does not match the numerator up to a constant factor. Instead, we will split the fraction into two simpler pieces. First, learn how each piece integrates.

You already know

Partial fractions express one rational function as a sum of simpler fractions on its original domain. Differentiating a logarithm gives $\frac{d}{dx}\ln|f(x)|=f'(x)/f(x)$ where $f(x)\ne0$ .

The reciprocal $1/(2x-3)$ gives a logarithm with coefficient $1/2$ . The squared reciprocal $(x+1)^{-2}$ instead uses the power rule, giving $-(x+1)^{-1}$ . The exponent determines which rule applies.

Logarithmic integration

On an interval where $f(x)\ne0$ ,

\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C

In the formula booklet.

For constants $a\ne0$ and $k$ , use the forms below on intervals where $ax+b\ne0$ . A repeated factor uses a power instead:

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

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

These two specialised forms are not separately printed: derive the logarithmic one from the supplied rule and learn the power-rule method. In both results the factor $1/a$ reverses the chain rule.

Worked example 1: a logarithm and a reciprocal

Question. Find $\int\left(\frac4{2x-3}-\frac3{(x+1)^2}\right)\,dx$ , for $x\ne3/2,-1$ .

Plan. Integrate the first term logarithmically and the repeated-factor term by the power rule.

\begin{aligned} \int\frac4{2x-3}\,dx&=2\ln|2x-3| && \text{divide by inside derivative }2 \\ \int-3(x+1)^{-2}\,dx&=3(x+1)^{-1} && \text{raise power to }-1 \\ \int\left(\frac4{2x-3}-\frac3{(x+1)^2}\right)\,dx&=2\ln|2x-3|+\frac3{x+1}+C && \text{combine} \end{aligned}

Why this step? Only a first power of a linear denominator gives a logarithm; the squared factor gives a reciprocal.

Answer. $2\ln|2x-3|+3/(x+1)+C$ , on an interval avoiding the two excluded values.

Check. Its derivative is $2\times2/(2x-3)-3/(x+1)^2$ , the original integrand.

Change only the slope of the first linear factor, from $2$ to $3$ . Its logarithm coefficient and excluded zero change; the other term stays the same.

Worked example 2: a changed inside derivative

Question. Find $\int\left(\frac4{3x-3}-\frac3{(x+1)^2}\right)\,dx$ , for $x\ne1,-1$ .

Plan. Adjust the logarithm's chain factor and keep the reciprocal term unchanged.

\begin{aligned} \int\frac4{3x-3}\,dx&=\frac43\ln|3x-3| && \text{inside derivative }3 \\ \int-3(x+1)^{-2}\,dx&=\frac3{x+1} && \text{same power integral} \\ \int\left(\frac4{3x-3}-\frac3{(x+1)^2}\right)\,dx&=\frac43\ln|3x-3|+\frac3{x+1}+C && \text{combine} \end{aligned}

Why this step? The new coefficient $4/3$ , multiplied by the logarithm's inside derivative $3$ , restores the numerator $4$ .

Answer. $\frac43\ln|3x-3|+3/(x+1)+C$ , on any interval avoiding $-1$ and $1$ .

Check. Differentiating gives $4/(3x-3)-3/(x+1)^2$ .

The absolute values allow a real logarithm on either side of each denominator zero. For instance, $\ln|3x-3|=\ln(3-3x)$ for $x<1$ . You still cannot cross $x=1$ ; constants of integration on separate intervals need not agree.

Watch out

Wrong: $\int(x+1)^{-2}\,dx=\ln|(x+1)^2|+C$ . That logarithm differentiates to $2/(x+1)$ . Use the power rule instead: $-(x+1)^{-1}+C$ .

Change only the slope of the first factor again in your turn.

Decomposing and integrating distinct linear factors

For two distinct linear factors, give each its own constant numerator. Multiplying through by the original denominator creates a polynomial identity. Coefficient comparison or suitable substitutions then finds the constants.

For example, $A/(x-1)+B/(x+2)$ combines to numerator $A(x+2)+B(x-1)$ . Setting $x=1$ in that identity removes the $B$ -term. This determines coefficients in a polynomial identity; it does not evaluate the original fraction at a forbidden point.

Worked example 3: solve the opening integral

Question. Find $\int\frac{5x+1}{(x-1)(x+2)}\,dx$ .

Plan. Find the two constant numerators, then integrate each simple fraction.

\begin{aligned} 5x+1&\equiv A(x+2)+B(x-1) && \text{equate numerators} \\ x=1&:\quad6=3A\Rightarrow A=2 && \text{remove }B \\ x=-2&:\quad-9=-3B\Rightarrow B=3 && \text{remove }A \\ \int\frac{5x+1}{(x-1)(x+2)}\,dx&=\int\left(\frac2{x-1}+\frac3{x+2}\right)\,dx && \text{split} \\ &=2\ln|x-1|+3\ln|x+2|+C && \text{integrate} \end{aligned}

Why this step? Each separated denominator is linear, so its antiderivative is now a logarithm with the correct weight.

Answer. $2\ln|x-1|+3\ln|x+2|+C$ , on an interval excluding $1$ and $-2$ .

Check. Recombining the differentiated terms gives numerator $2(x+2)+3(x-1)=5x+1$ .

Change only the constant term in the numerator, from $1$ to $4$ . The denominator and exclusions stay the same, but the logarithm weights change.

Worked example 4: change the numerator weights

Question. Find $\int\frac{5x+4}{(x-1)(x+2)}\,dx$ .

Plan. Use the same two simple fractions and determine their new constants.

\begin{aligned} 5x+4&\equiv A(x+2)+B(x-1) && \text{numerator identity} \\ x=1&:\quad9=3A\Rightarrow A=3 && \text{find }A \\ x=-2&:\quad-6=-3B\Rightarrow B=2 && \text{find }B \\ \int\frac{5x+4}{(x-1)(x+2)}\,dx&=3\ln|x-1|+2\ln|x+2|+C && \text{integrate the fractions} \end{aligned}

Why this step? Changing the numerator changes the coefficients; the factorisation of the denominator alone does not determine them.

Answer. $3\ln|x-1|+2\ln|x+2|+C$ , on an interval avoiding the original poles.

Check. $3(x+2)+2(x-1)=5x+4$ , so differentiation and recombination recover the original fraction.

The logarithm of the whole denominator would differentiate to $(2x+1)/[(x-1)(x+2)]$ . Neither example has that numerator up to a constant factor. You may combine the correctly weighted logarithms afterwards, but cannot guess their weights from the denominator. These decompositions are derived by algebra; they are not in the formula booklet.

Watch out

Wrong: $\int\frac{5x+1}{(x-1)(x+2)}\,dx=\ln|(x-1)(x+2)|+C$ . Its derivative has numerator $2x+1$ . Correctly decompose first, giving the weights $2$ and $3$ .

Change only the numerator's constant again and check your coefficients by recombining before integrating.

Repeated factors and preparing the fraction

A squared linear factor needs terms for both its first and second powers. For $(x+1)^2(x+2)$ , use $A/(x+1)+B/(x+1)^2+D/(x+2)$ . Omitting the first-power term usually leaves too few constants to match the numerator.

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H6 - Integration using partial fractions

Integrating individual linear-denominator terms

Decomposing and integrating distinct linear factors

Repeated factors and preparing the fraction

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