CP 5.4 - Partial fractions with quadratic factors | Edexcel A-Level Further Mathematics

A quadratic denominator can produce both a logarithm and an inverse tangent. You will choose the correct partial-fraction numerators, integrate the resulting terms, and handle improper fractions and exact definite integrals.

The numerator over a quadratic factor

How can a fraction made entirely from polynomials have an angle in its antiderivative? A factor such as $x^2+1$ is the clue: its reciprocal integrates to arctangent. First you need a partial-fraction form that does not throw away part of the numerator.

You already know

A-level Mathematics partial fractions split a rational expression into simpler fractions. Multiply by the common denominator and compare coefficients or substitute convenient values. You can integrate $f'(x)/f(x)$ as $\ln|f(x)|$ , and the previous lesson gives $\int\mathrm{d}x/(a^2+x^2)$ .

For a linear denominator, the numerator is a constant. For an irreducible quadratic denominator, it must be a general linear expression. “Irreducible” here means that the quadratic has no real linear factors: $x^2+1$ is irreducible, while $x^2-1=(x-1)(x+1)$ is not. For $ax^2+c$ , positive $a$ and $c$ give an irreducible quadratic; an overall negative factor can be taken outside.

A linear factor and an irreducible quadratic

For a proper fraction, with numerator degree below $3$ ,

\frac{P(x)}{(x-r)(ax^2+c)} =\frac{A}{x-r}+\frac{Bx+C}{ax^2+c}

where the quadratic is irreducible and the denominators are non-zero. A coefficient may turn out to be zero, but do not assume that before solving.

Not in the formula booklet: learn it.

The quadratic numerator has two coefficients because its degree can be up to one less than the denominator's degree. Using just a constant would wrongly rule out an $x$ -term that the original expression may need.

Worked example 1: finding three coefficients

Question. Express $\frac{3x^2+2x+3}{(x+1)(x^2+1)}$ in partial fractions.

Plan. Use $A/(x+1)+(Bx+C)/(x^2+1)$ , clear denominators, then compare coefficients.

\begin{aligned} 3x^2+2x+3&=A(x^2+1)+(Bx+C)(x+1) && \text{clear denominators}\\ &=(A+B)x^2+(B+C)x+(A+C) && \text{expand} \end{aligned}

Thus $A+B=3$ , $B+C=2$ , $A+C=3$ . The first and third give $B=C$ , hence $B=C=1$ , and $A=2$ .

Why this form? The general linear numerator allows the coefficients of all three powers to match.

Answer.

\frac2{x+1}+\frac{x+1}{x^2+1}

Check. The combined numerator is $2(x^2+1)+(x+1)^2=3x^2+2x+3$ .

Worked example 2 changes only the coefficient of $x$ in the original numerator. The required fraction form stays exactly the same.

Worked example 2: the same denominator

Question. Express $\frac{3x^2+4x+3}{(x+1)(x^2+1)}$ in partial fractions.

Plan. Use the same denominator pattern, but solve the new coefficient equations.

\begin{aligned} A+B&=3 && x^2\text{ coefficients}\\ B+C&=4 && x\text{ coefficients}\\ A+C&=3 && \text{constant terms} \end{aligned}

The first and third again give $B=C$ , so $B=C=2$ , then $A=1$ .

Why solve again? Partial-fraction coefficients depend on the numerator as well as the denominator.

Answer.

\frac1{x+1}+\frac{2x+2}{x^2+1}

Check. $(x^2+1)+(2x+2)(x+1)=3x^2+4x+3$ .

You may substitute $x=-1$ into the cleared polynomial identity, even though the original rational expression is undefined there. You are evaluating an identity between polynomials, not evaluating the original fraction at its pole.

Watch out

Wrong: $A/(x+1)+B/(x^2+1)$ as the general form. Right: use $(Bx+C)/(x^2+1)$ . A constant numerator is a possible result, not a valid assumption for every quadratic factor.

Change the numerator once more in your own decomposition.

Separating logarithm and arctangent terms

The two terms in a quadratic numerator integrate differently. Split them before choosing a rule: the $x$ -term can match the denominator's derivative, while the constant term gives an arctangent.

For $a>0$ , $c>0$ , a term $px/(ax^2+c)$ integrates to $\frac{p}{2a}\ln(ax^2+c)$ . This follows from the derivative of the denominator, $2ax$ , and is not listed separately in the formula booklet. No absolute value is needed here because $ax^2+c>0$ . A linear denominator still needs $\ln|x-r|$ unless the interval fixes its sign.

For the constant numerator $q$ , factor out $c$ and scale the variable by $\sqrt{a/c}$ . The supplied arctangent formula then gives

\int\frac{q}{ax^2+c}\,\mathrm{d}x =\frac{q}{\sqrt{ac}}\arctan\left(x\sqrt{\frac ac}\right)+C

This general combination is not printed separately in the formula booklet: derive it from the supplied standard integral. Together, the two numerator pieces handle any positive quadratic of the required form, not just one with constant term $1$ .

Worked example 3: integrating a decomposition

Question. Find

\int\left(\frac2{x+1}+\frac{x+1}{x^2+1}\right)\mathrm{d}x

Plan. Split the quadratic numerator into an $x$ -term and a constant term, then integrate each type.

\begin{aligned} I&=2\int\frac{\mathrm{d}x}{x+1} +\int\frac{x}{x^2+1}\,\mathrm{d}x +\int\frac{\mathrm{d}x}{x^2+1} && \text{split the numerator}\\ &=2\ln|x+1|+\frac12\ln(x^2+1)+\arctan x+C && \text{log, log, angle} \end{aligned}

Why the factor $\frac12$ ? The derivative of $x^2+1$ is $2x$ , twice the numerator of that term.

Answer. $2\ln|x+1|+\frac12\ln(x^2+1)+\arctan x+C$ , on any interval avoiding $x=-1$ .

Check. Differentiation gives $2/(x+1)+x/(x^2+1)+1/(x^2+1)$ , the original expression.

Worked example 4 changes only the coefficient of $x^2$ in the quadratic denominator. Both the logarithmic and arctangent scale factors must respond.

Worked example 4: a non-unit quadratic coefficient

Question. Find

\int\left(\frac2{x+1}+\frac{x+1}{4x^2+1}\right)\mathrm{d}x

Plan. Split the numerator again, match $8x$ for the logarithm, and use $u=2x$ for the constant term.

\begin{aligned} I&=2\ln|x+1|+\frac18\ln(4x^2+1) +\int\frac{\mathrm{d}x}{1+(2x)^2} && \text{integrate the log terms}\\ &=2\ln|x+1|+\frac18\ln(4x^2+1) +\frac12\arctan(2x)+C && \text{scale the arctangent} \end{aligned}

Why two different factors? The logarithm uses the derivative $8x$ ; the arctangent uses a linear variable scale $2x$ . They are not the same adjustment.

Answer. $2\ln|x+1|+\frac18\ln(4x^2+1)+\frac12\arctan(2x)+C$ .

Check. The last two derivatives are $x/(4x^2+1)$ and $1/(4x^2+1)$ , respectively.

This is the source of the angle in the opening puzzle: the constant part over a positive quadratic uses the supplied arctangent integral. The numerator's $x$ -part produces a logarithm alongside it.

Watch out

Wrong: integrate $(x+1)/(x^2+1)$ as $\ln(x^2+1)$ . Right: split it; only the $x$ -part is proportional to the denominator's derivative. The constant part gives $\arctan x$ .

Keep those two scale adjustments separate in the next integral.

Dividing before splitting an improper fraction

If the numerator's degree is at least the denominator's degree, ordinary proper partial fractions cannot represent the whole expression. Divide first, or include a polynomial term in your proposed form.

Proper and improper rational fractions

A rational fraction is proper when its numerator has lower degree than its denominator. It is improper when the numerator has equal or greater degree. For example, a cubic numerator over $(x+1)(x^2+1)$ requires a constant polynomial part as well as proper fractions. If the numerator degree were still higher, the polynomial part could have positive degree.

The denominator below expands to $x^3+x^2+x+1$ . Subtracting a multiple of that polynomial leaves a remainder of degree at most two, ready for the form already taught.

Worked example 5: an improper integrand

Question. Express

\frac{2x^3+6x^2+5x+7}{(x+1)(x^2+1)}

in partial fractions, and hence integrate it.

Plan. Divide to obtain the constant part, split the remainder, then integrate all terms.

\begin{aligned} 2x^3+6x^2+5x+7 &=2(x^3+x^2+x+1)+(4x^2+3x+5) && \text{division with remainder}\\ 4x^2+3x+5 &=3(x^2+1)+(x+2)(x+1) && \text{decompose the remainder} \end{aligned}

Thus the expression is $2+\frac3{x+1}+\frac{x+2}{x^2+1}$ .

Why keep the $2$ ? Every proper fraction tends to zero as $|x|$ grows, while the original expression tends to $2$ . Omitting the constant cannot possibly give an identity.

Answer. The partial-fraction form is

2+\frac3{x+1}+\frac{x+2}{x^2+1}

and its integral is

2x+3\ln|x+1|+\frac12\ln(x^2+1)+2\arctan x+C

Check. Differentiation restores all four terms, including the constant $2$ . Combining the proper fractions restores the remainder $4x^2+3x+5$ .

Worked example 6 changes the numerator by subtracting one copy of the denominator. Only the polynomial part changes; the proper remainder is unchanged.

Worked example 6: changing the polynomial part

Question. Find

\int\frac{x^3+5x^2+4x+6}{(x+1)(x^2+1)}\,\mathrm{d}x

Plan. Divide first, recognise the same remainder, then integrate the new constant part.

\begin{aligned} x^3+5x^2+4x+6 &=(x^3+x^2+x+1)+(4x^2+3x+5) && \text{division}\\ \frac{x^3+5x^2+4x+6}{(x+1)(x^2+1)} &=1+\frac3{x+1}+\frac{x+2}{x^2+1} && \text{split the remainder} \end{aligned}

Why this step? The quotient is now $1$ , so the polynomial part of the antiderivative is $x$ , rather than $2x$ .

Answer.

x+3\ln|x+1|+\frac12\ln(x^2+1)+2\arctan x+C

Check. Subtracting this answer from Example 5's answer gives $x$ , whose derivative is the one unit removed from the integrand.

You can also start with $k+A/(x+1)+(Bx+C)/(x^2+1)$ and compare coefficients directly. Division usually makes the size of the polynomial part easier to see. These decomposition methods are not in the formula booklet.

Watch out

Wrong: give only $\frac3{x+1}+\frac{x+2}{x^2+1}$ for the fraction in Worked example 5. That expression tends to zero as $|x|$ grows, while the original fraction tends to $2$ . Right: retain the quotient, giving $2+\frac3{x+1}+\frac{x+2}{x^2+1}$ .

The polynomial term belongs in the assembled answer as well as in the division working.

Exam tip

When asked to express a fraction in partial fractions, finish by writing the complete expression. Listing coefficient values without assembling the constant and fractional terms can leave the requested answer unfinished.

The next numerator adds one copy of the denominator to Example 5. Identify the new constant part and keep the remainder.

Exact definite integrals with logarithms and angles

After integration, keep the logarithms and inverse angles exact until the limits have been substituted. Many attractive exact answers arise because the arctangent arguments are standard tangent values.

First check the interval contains no denominator zero. The functions below are continuous for non-negative $x$ , so ordinary evaluation of an antiderivative is valid there.

Worked example 7: a rational integral producing pi

Question. Evaluate exactly

\int_0^1\frac{3x^2+2x+3}{(x+1)(x^2+1)}\,\mathrm{d}x

Plan. Use the partial fractions from Example 1, integrate, then substitute the limits without rounding the angle.

\begin{aligned} I&=\left[2\ln(x+1)+\frac12\ln(x^2+1)+\arctan x\right]_0^1\\ &=2\ln2+\frac12\ln2+\frac\pi4 && \arctan1=\pi/4\\ &=\frac52\ln2+\frac\pi4 && \text{collect logarithms} \end{aligned}

Why no lower-limit term remains? At $x=0$ , both logarithms are $\ln1=0$ and $\arctan0=0$ .

Answer. $\frac52\ln2+\frac\pi4$ .

Check. The integral is positive. Numerically it is approximately $2.52$ , consistent with a positive integrand of modest size on a unit interval.

Worked example 8 changes only the upper endpoint to $\sqrt{3}$ , bringing in a different exact angle and a different logarithmic simplification.

Worked example 8: combining exact logarithms

Question. Evaluate exactly

\int_0^{\sqrt{3}}\frac{3x^2+2x+3}{(x+1)(x^2+1)}\,\mathrm{d}x

Plan. Use the same antiderivative, substitute $\sqrt{3}$ , and combine the logarithms.

\begin{aligned} I&=2\ln(1+\sqrt{3})+\frac12\ln4+\frac\pi3 && \arctan\sqrt{3}=\pi/3\\ &=\ln\bigl(2(1+\sqrt{3})^2\bigr)+\frac\pi3 && \text{logarithm laws}\\ &=\ln(8+4\sqrt{3})+\frac\pi3 && \text{expand the square} \end{aligned}

Why combine after substitution? Each logarithm now has a positive numerical argument, so the logarithm laws can be applied directly without a domain ambiguity.

Answer. $\ln(8+4\sqrt{3})+\frac\pi3$ .

Check. The upper endpoint is larger and the integrand is positive, so the answer must exceed Example 7's value. Its approximate value, $3.75$ , does.

There is a neat mathematical surprise here: a rational integrand with integer coefficients produces a sum involving both $\ln2$ and $\pi$ . The logarithm comes from the linear and derivative-matching terms; the circle constant arrives through the inverse angle. The denominator quietly determines which kinds of function appear.

Watch out

Wrong: replace $\arctan\sqrt{3}$ by $60$ in an exact integral. Right: use $\pi/3$ . Calculus uses radians, and a decimal angle would also lose the exact form.

Change the lower endpoint in the final check. This time it contributes non-zero logarithm and angle terms.

Quick Recap

Use a general linear numerator over an irreducible quadratic factor.
Split that numerator: the derivative-matching term gives a logarithm, while the constant term gives an arctangent.
Divide an improper fraction first and retain its polynomial part when integrating.
For definite integrals, check the interval, substitute both limits and keep angles in exact radians.

The opening polynomial fraction produces an angle because its positive quadratic denominator contains an arctangent integral.

Last Accessed

Core Pure Mathematics

Further Pure Mathematics 1

Further Pure Mathematics 2

Further Statistics 1

Further Statistics 2

Further Mechanics 1

Further Mechanics 2

Decision Mathematics 1

Decision Mathematics 2

CP 5.4 - Partial fractions with quadratic factors

The numerator over a quadratic factor

Separating logarithm and arctangent terms

Dividing before splitting an improper fraction

Exact definite integrals with logarithms and angles

Quick Recap

Learn faster with Nova