Pc4 - Integrating with partial fractions | OCR A-Level Further Mathematics B (MEI)

Extend partial-fraction integration to a denominator containing one linear factor and a quadratic factor of the form $ax^2+c$ . Pc4 combines logarithmic integrals with the inverse-tangent integrals from the previous lesson, keeping decomposition, integration and endpoint evaluation as distinct steps.

A quadratic factor needs a linear numerator

How can the rational integral

\int_0^2\frac{2x^2+3x+6}{(x+1)(x^2+4)}\,dx

produce an exact answer involving $\pi$ ? Its denominator hides a sum of squares. Partial fractions will expose that piece, alongside two logarithmic pieces.

You already know

For a proper rational fraction with linear factors, partial fractions separate the factors before integration. If the numerator's degree is at least the denominator's degree, use polynomial division first. Also, $\int dx/(x-h)=\ln|x-h|+C$ .

The numerator above a factor must allow every polynomial of degree below that factor's degree. A linear factor needs a constant numerator. An irreducible quadratic needs a general linear numerator, not just a constant.

A linear factor and an irreducible quadratic

For a proper rational fraction with denominator $(x-h)(ax^2+c)$ , where $ax^2+c$ has no real roots, use

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

Multiply by the common denominator, then use substitution and equating coefficients to find $A,B,C$ .

Not in the formula booklet: learn it.

For positive $a,c$ , the quadratic is always positive and cannot be factored into real linear terms. If $a$ and $c$ have opposite signs, it factors into real linear terms instead; use your earlier linear-factor methods. For example, $x^2-4=(x-2)(x+2)$ . If $c=0$ , treat the repeated linear factor explicitly.

Worked example 1: choosing enough numerator terms

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

Plan. Use a constant above $x+1$ , a linear expression above $x^2+1$ , then form a polynomial identity.

\begin{aligned} \frac{x+3}{(x+1)(x^2+1)}&=\frac A{x+1}+\frac{Bx+C}{x^2+1} \\ x+3&=A(x^2+1)+(Bx+C)(x+1) && \text{clear denominators} \\ x=-1&\ \Longrightarrow\ 2=2A,\quad A=1 && \text{remove the second product} \\ x^2:\quad A+B&=0,\quad B=-1 && \text{equate coefficients} \\ x:\quad B+C&=1,\quad C=2 && \text{equate coefficients} \end{aligned}

Why this step? Substituting $x=-1$ into the polynomial identity is valid, even though it is excluded from the original fraction. The identity has no denominator after it has been cleared.

Answer. $\dfrac1{x+1}+\dfrac{-x+2}{x^2+1}$ , for $x\ne-1$ .

Check. Recombining gives numerator $(x^2+1)+(-x+2)(x+1)=x+3$ .

Change only the constant term in the original numerator, from $3$ to $5$ . Keep the same decomposition structure; the coefficients, rather than the allowed terms, will change.

Worked example 2: a changed numerator

Question. Express $\dfrac{x+5}{(x+1)(x^2+1)}$ in partial fractions.

Plan. Form the same identity, use $x=-1$ , then equate coefficients.

\begin{aligned} x+5&=A(x^2+1)+(Bx+C)(x+1) && \text{clear denominators} \\ x=-1&\ \Longrightarrow\ 4=2A,\quad A=2 \\ x^2:\quad A+B&=0,\quad B=-2 \\ x:\quad B+C&=1,\quad C=3 \end{aligned}

Why this step? A constant-only numerator over the quadratic would force $B=0$ , making the $x^2$ coefficient impossible to cancel here.

Answer. $\dfrac2{x+1}+\dfrac{-2x+3}{x^2+1}$ , for $x\ne-1$ .

Check. The combined numerator is $2(x^2+1)+(-2x+3)(x+1)=x+5$ .

The general numerator is a way to avoid ruling out the correct answer before starting. A coefficient may turn out to be zero, but that is a conclusion of the identity.

Watch out

Wrong starting form: $A/(x+1)+C/(x^2+1)$ .

Right starting form: $A/(x+1)+(Bx+C)/(x^2+1)$ . Omitting $Bx$ loses the term needed to cancel unwanted quadratic coefficients.

Another constant change provides a quick check of the same coefficient method.

Split the quadratic term into a logarithm and an angle

After decomposition, a term $(Bx+C)/(ax^2+c)$ has two different integration routes. The $Bx$ numerator is proportional to the denominator's derivative, so it gives a logarithm. The constant numerator gives an inverse tangent when the quadratic is positive.

For instance, $d(x^2+4)/dx=2x$ , so an $x/(x^2+4)$ term gives $\frac12\ln(x^2+4)$ . Meanwhile $1/(x^2+4)$ gives $\frac12\arctan(x/2)$ .

Two pieces of a quadratic fraction

For $a>0,c>0$ ,

\begin{aligned} \int\frac{Bx}{ax^2+c}\,dx&=\frac{B}{2a}\ln(ax^2+c)+K \\ \int\frac{C}{ax^2+c}\,dx &=\frac{C}{\sqrt{ac}}\arctan\left(x\sqrt{\frac ac}\right)+K \end{aligned}

The logarithmic rule follows by reverse differentiation. The second line follows by scaling the standard inverse-tangent integral, which is supplied.

Not in the formula booklet: learn it.

Use a single arbitrary constant after adding indefinite integrals. The different constants in intermediate pieces combine into one constant; they are not separate requirements.

Worked example 3: two numerator terms, two methods

Question. Find $\int\dfrac{3x+4}{x^2+4}\,dx$ .

Plan. Separate the $x$ term from the constant term, then use a logarithm and an inverse tangent.

\begin{aligned} I&=3\int\frac{x}{x^2+4}\,dx+4\int\frac1{x^2+4}\,dx && \text{split numerator} \\ &=\frac32\ln(x^2+4)+4\left(\frac12\arctan\frac x2\right)+K && \text{two standard routes} \\ &=\frac32\ln(x^2+4)+2\arctan\frac x2+K \end{aligned}

Why this step? A denominator is not enough to choose an integration rule: its numerator decides whether the piece is logarithmic or inverse-trigonometric.

Answer. $\frac32\ln(x^2+4)+2\arctan(x/2)+K$ .

Check. Differentiation gives $3x/(x^2+4)+4/(x^2+4)$ , reconstructing the original numerator.

Change only the coefficient of $x^2$ in the denominator, from $1$ to $4$ . Both integration coefficients must respond to that change.

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Pc4 - Integrating with partial fractions

A quadratic factor needs a linear numerator

Split the quadratic term into a logarithm and an angle

Continue this lesson.

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