IB Analysis and Approaches HL · Integral calculus · Section 4 of 13

Integrating rational functions

3 worked examples · By Álvaro López

Key idea

A rational function is a quotient of polynomials, $\frac{P ( x )}{Q ( x )}$ . To integrate it, divide if necessary and split it into partial fractions, each of which has an immediate antiderivative (a logarithm, an arctangent or a power). This is the method that earns the most marks in the long integration questions at HL.

Preliminary step: division

If the degree of $P$ is greater than or equal to that of $Q$ , divide: $\frac{P ( x )}{Q ( x )} = C (x) + \frac{R ( x )}{Q ( x )},$ where $C$ is a polynomial (easy to integrate) and the remainder $R$ has a smaller degree than $Q$ . The problem then reduces to the case where the numerator is "smaller" than the denominator.

Partial fraction decomposition

With $degree P < degree Q$ , factorise $Q$ and write one fraction for each factor:

A factor repeated $k$ times contributes $k$ fractions, with all powers from 1 up to $k$ . An irreducible quadratic factor contributes a numerator of degree 1, not a constant.

Fractions with an irreducible quadratic are integrated by splitting $\frac{A x + B}{\dots}$ into a part that is the derivative of the denominator (giving a logarithm) and a constant part (giving an arctangent).

When there are no real roots: completing the square

For $\int \frac{d x}{x ^{2} + b x + c}$ with a negative discriminant, complete the square: $x^{2} + b x + c = (x + h)^{2} + k^{2}$ . With the substitution $u = x + h$ you obtain $\frac{1}{k} arctan \frac{x + h}{k}$ .

Worked examples

Example 1 Basic

Two distinct roots

Calculate $\int \frac{3 x + 5}{( x + 1 ) ( x + 2 )} d x$ , for $x > - 1$ .

Solution

Decompose: $\frac{3 x + 5}{( x + 1 ) ( x + 2 )} = \frac{A}{x + 1} + \frac{B}{x + 2} \Rightarrow 3 x + 5 = A (x + 2) + B (x + 1) .$ With $x = - 1$ : $2 = A \cdot 1$ , so $A = 2$ . With $x = - 2$ : $- 1 = B \cdot (- 1)$ , so $B = 1$ . $\int \frac{3 x + 5}{( x + 1 ) ( x + 2 )} d x = \int (\frac{2}{x + 1} + \frac{1}{x + 2}) d x = 2 ln (x + 1) + ln (x + 2) + C .$

Example 2 Medium

Prior division, a repeated root and completing the square

Calculate:

a) $\int \frac{x ^{2} + 3}{x ^{2} - 1} d x$ , for $x > 1$ .

b) $\int \frac{d x}{x ( x + 1 ) ^{2}}$ , for $x > 0$ .

c) $\int \frac{d x}{x ^{2} + 2 x + 5}$ .

Solution

a) Same degree, so divide: $x^{2} + 3 = 1 \cdot (x^{2} - 1) + 4$ , so $\frac{x ^{2} + 3}{x ^{2} - 1} = 1 + \frac{4}{( x - 1 ) ( x + 1 )} = 1 + \frac{2}{x - 1} - \frac{2}{x + 1} .$ (The two fractions: $4 = A (x + 1) + B (x - 1)$ ; with $x = 1$ this gives $A = 2$ , and with $x = - 1$ it gives $B = - 2$ .) Then $\int \frac{x ^{2} + 3}{x ^{2} - 1} d x = x + 2 ln (x - 1) - 2 ln (x + 1) + C .$

b) The factor $(x + 1)$ is repeated: $\frac{1}{x ( x + 1 ) ^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{( x + 1 ) ^{2}} \Rightarrow 1 = A (x + 1)^{2} + B x (x + 1) + C x .$ With $x = 0$ : $A = 1$ . With $x = - 1$ : $1 = - C$ , so $C = - 1$ . With $x = 1$ : $1 = 4 A + 2 B + C = 4 + 2 B - 1$ , so $B = - 1$ . Then $\int \frac{d x}{x ( x + 1 ) ^{2}} = ln x - ln (x + 1) + \frac{1}{x + 1} + C .$

c) The discriminant is $4 - 20 < 0$ . Complete the square: $x^{2} + 2 x + 5 = (x + 1)^{2} + 4$ . With $u = x + 1$ : $\int \frac{d x}{( x + 1 ) ^{2} + 4} = \frac{1}{2} arctan \frac{x + 1}{2} + C .$

Example 3 Hard

An irreducible quadratic factor with limits

Let $f (x) = \frac{2 x ^{2} + x + 2}{x ( x ^{2} + 1 )}$ .

a) Find the constants $A$ , $B$ and $C$ such that $f (x) = \frac{A}{x} + \frac{B x + C}{x ^{2} + 1}$ .

b) Calculate the exact value of $\int_{1}^{2} f (x) d x$ .

Solution

a) Multiplying by $x (x^{2} + 1)$ : $2 x^{2} + x + 2 = A (x^{2} + 1) + (B x + C) x .$ With $x = 0$ : $2 = A$ . Equating coefficients of $x^{2}$ : $2 = A + B$ , so $B = 0$ . From $x$ : $1 = C$ . Then $f (x) = \frac{2}{x} + \frac{1}{x ^{2} + 1} .$

b) Integrate each fraction: $\int \frac{2}{x} d x = 2 ln x$ and $\int \frac{d x}{x ^{2} + 1} = arctan x$ . $\int_{1}^{2} f (x) d x = [2 ln x + arctan x]_{1}^{2} = 2 ln 2 + arctan 2 - \frac{π}{4} .$ (Approximately $1.708$ .)

In the exam: in a) marks are given for the form $\frac{B x + C}{x ^{2} + 1}$ of the second term; if you put a constant over the quadratic you will not be able to match the numerator. In b) leave the exact answer.

Common mistakes

Summary

If $degree P \geq degree Q$ , divide first: $\frac{P}{Q} = C + \frac{R}{Q}$ .
Factorise $Q$ and decompose: one fraction per linear factor, one per power of a repeated factor, and $\frac{A x + B}{quadratic}$ for an irreducible quadratic.
Find the letters by substituting the roots of $Q$ and, if any are missing, other values or the coefficients.
Resulting integrals: $\frac{A}{x - a} \to A ln ∣ x - a ∣$ ; $\frac{B}{( x - a ) ^{2}} \to - \frac{B}{x - a}$ ; quadratics $\to arctan$ after completing the square.
For definite integrals, evaluate the limits in the exact result using $ln$ and $arctan$ .

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Factor of $Q (x)$	Fractions it contributes
$(x - a)$	$\frac{A}{x - a}$
$(x - a)^{2}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$
$(x - a)^{3}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{( x - a ) ^{3}}$
$(p x^{2} + q x + r)$ , with no real roots	$\frac{A x + B}{p x ^{2} + q x + r}$