3.5 - Integrating partial fractions | Cambridge A-Level Mathematics

A rational function can hide several familiar integrals inside one fraction. In this part of 3.5, you will decompose it first, then recognise logarithmic, power and inverse-tangent terms. You will also evaluate exact definite integrals and check that their intervals do not cross an undefined input.

Linear denominators

How can you integrate $(5x+7)/[(x+1)(2x+3)]$ ? Its numerator is not a constant multiple of the derivative of its whole denominator. Splitting it as $2/(x+1)+1/(2x+3)$ reveals two logarithmic integrals instead.

You already know

For a proper rational fraction with distinct linear factors, put a constant numerator over each factor. Clear the denominator and determine those constants by substitution or coefficient comparison. Recombining checks the identity.

Once the fractions are separate, integrate each one. A factor's coefficient of $x$ matters just as much as its constant numerator.

Integrating a linear-denominator term

For $a\ne0$ , on an interval where $ax+b\ne0$ ,

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

The basic logarithmic integral is in the formula booklet. This scaled form follows by reverse differentiation; it is not listed separately.

Absolute values allow a negative denominator, but never a zero denominator. An indefinite answer applies separately on each interval between the original denominator's zeros.

Worked example 1: two logarithmic terms

Question. Find $\displaystyle\int\frac{5x+7}{(x+1)(2x+3)}\,\mathrm dx$ .

Plan. Find the two partial fractions, then divide each logarithm coefficient by the derivative of its linear argument.

Write

5x+7=A(2x+3)+B(x+1).

At $x=-1$ , $2=A$ . At $x=-3/2$ , $-1/2=-B/2$ , so $B=1$ . Hence

\begin{aligned} \int\frac{5x+7}{(x+1)(2x+3)}\,\mathrm dx &=\int\left(\frac2{x+1}+\frac1{2x+3}\right)\,\mathrm dx\\ &=2\ln|x+1|+\frac12\ln|2x+3|+C \end{aligned}

Answer. $2\ln|x+1|+\frac12\ln|2x+3|+C$ , on an interval excluding $-1$ and $-3/2$ .

Check. Differentiation gives $2/(x+1)+1/(2x+3)$ . Its combined numerator is $2(2x+3)+(x+1)=5x+7$ .

Now change the second factor from $2x+3$ to $3-2x$ , keeping the partial-fraction numerators $2$ and $1$ . Its derivative changes sign, so the second logarithmic coefficient must change sign too.

Worked example 2: a negative inner derivative

Question. Find $\displaystyle\int\frac{7-3x}{(x+1)(3-2x)}\,\mathrm dx$ .

Plan. Decompose as before, retaining the negative derivative of $3-2x$ when integrating.

Clearing the proposed fractions gives

7-3x=A(3-2x)+B(x+1).

At $x=-1$ , $10=5A$ , so $A=2$ . At $x=3/2$ , $5/2=5B/2$ , so $B=1$ . Thus

\int\left(\frac2{x+1}+\frac1{3-2x}\right)\,\mathrm dx =2\ln|x+1|-\frac12\ln|3-2x|+C.

Answer. $2\ln|x+1|-\frac12\ln|3-2x|+C$ , on an interval excluding $-1$ and $3/2$ .

Check. The second term differentiates to $(-1/2)(-2)/(3-2x)=1/(3-2x)$ . Recombining the derivatives gives numerator $2(3-2x)+(x+1)=7-3x$ .

For three distinct linear factors, the same process produces three logarithmic terms. If numerator and denominator have equal degree, first remove the constant quotient and integrate that constant too. The allowed decomposition forms do not extend here to a numerator of higher degree than its denominator.

Watch out

A partial-fraction coefficient is not automatically the logarithm's coefficient. The term $1/(3-2x)$ integrates to $-\frac12\ln|3-2x|$ , not $\ln|3-2x|$ . Differentiate the proposed logarithm before moving on.

Use the same sign check with a different decreasing denominator.

Repeated and quadratic denominators

A squared linear denominator produces a power integral, whereas an irreducible quadratic may produce both a logarithm and an inverse tangent. Decide from the separated terms, rather than treating every denominator as a logarithm.

For a repeated factor $(cx+d)^2$ , retain both terms $B/(cx+d)$ and $D/(cx+d)^2$ . The second is $D(cx+d)^{-2}$ , so the power rule gives $-D/[c(cx+d)]$ . It does not give a logarithm.

For an irreducible quadratic $cx^2+d$ , its partial-fraction numerator is linear. Split $(Bx+E)/(cx^2+d)$ into an $x$ -numerator term and a constant-numerator term. The first matches the denominator's derivative; the second can be scaled to $1/(x^2+a^2)$ .

The quadratic split

For $a>0$ ,

\int\frac{Bx+E}{x^2+a^2}\,\mathrm dx =\frac B2\ln(x^2+a^2)+\frac E a\tan^{-1}\!\left(\frac xa\right)+C.

The logarithmic-derivative rule and $\int 1/(x^2+a^2)\,\mathrm dx$ are in the formula booklet. This combined split is not listed separately: reconstruct it from those two results.

The quadratic is always positive, so the displayed logarithm needs no absolute values. If the quadratic is $cx^2+d$ , first factor out $c$ . For example, $1/(2x^2+1)=\frac12/[x^2+(1/\sqrt2)^2]$ , which integrates to $\tan^{-1}(\sqrt2x)/\sqrt2$ . Work in radians.

Worked example 3: a repeated linear factor

Question. Find $\displaystyle\int\frac{3x^2+14x+13}{(x+1)(x+2)^2}\,\mathrm dx$ .

Plan. Include both powers of $x+2$ ; integrate the first powers as logarithms and the square by the power rule.

The cleared identity is

3x^2+14x+13=A(x+2)^2+B(x+1)(x+2)+D(x+1).

At $x=-1$ , $2=A$ . At $x=-2$ , $-3=-D$ , so $D=3$ . Squared coefficients give $3=A+B$ , so $B=1$ . Therefore

\begin{aligned} I&=\int\left(\frac2{x+1}+\frac1{x+2}+\frac3{(x+2)^2}\right)\,\mathrm dx\\ &=2\ln|x+1|+\ln|x+2|-\frac3{x+2}+C \end{aligned}

Answer. $2\ln|x+1|+\ln|x+2|-3/(x+2)+C$ , away from $x=-1,-2$ .

Check. Differentiating the last nonconstant term gives $3/(x+2)^2$ . Recombining all three derivatives gives numerator $2(x+2)^2+(x+1)(x+2)+3(x+1)=3x^2+14x+13$ .

Change the repeated factor to an irreducible quadratic. The decomposition still has three adjustable constants, but its final term now involves inverse tangent rather than a negative power.

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3.5 - Integrating partial fractions

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