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Cambridge International AS & A Level · Topic 3.5
A-Level Maths 9709 / Topic 3.5 / AS & A Level

Split it, then log it.

Paper 3 only. When the top of a fraction is a multiple of the derivative of the bottom, the integral is a logarithm. When it is not, split the fraction into partial fractions first: each simple piece then integrates to a log, a power or an inverse tan.

Topic 3.5All integration lessons →
The key idea

\(\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c\). For any other proper rational function, write it in partial fractions; then \(\dfrac{A}{ax+b} \to \dfrac{A}{a}\ln|ax+b|\), \(\dfrac{B}{(ax+b)^2} \to -\dfrac{B}{a(ax+b)}\), and \(\dfrac{Cx+D}{x^2+k^2}\) splits into a log and an inverse tan.

x y 123 01 y = 2/(x + 1) + 1/(2x + 1) area = 2 ln 2 + (1/2) ln 3 ≈ 1.94
Fig. 1: y = (5x + 3)/((x + 1)(2x + 1)) split into partial fractions as 2/(x + 1) + 1/(2x + 1); each piece integrates to a logarithm, so the shaded area from x = 0 to x = 1 is 2 ln 2 + (1/2) ln 3, about 1.94
Section 01

Top is the derivative of the bottom: a logarithm.

Differentiate \(\ln|f(x)|\) with the chain rule and you get \(\dfrac{f'(x)}{f(x)}\). So before anything else, check whether the numerator is a constant multiple of the derivative of the denominator. If it is, the answer is a log straight away.

Worked example Fix the constant
Find the exact value of \(\displaystyle\int_0^2 \dfrac{x}{x^2+4}\,dx\).
  1. The derivative of \(x^2 + 4\) is \(2x\). The top is \(x = \dfrac{1}{2}(2x)\).
  2. So \(\displaystyle\int \dfrac{x}{x^2+4}\,dx = \dfrac{1}{2}\ln(x^2+4)\) (no modulus: \(x^2 + 4 > 0\)).
  3. Limits: \(\dfrac{1}{2}\ln 8 - \dfrac{1}{2}\ln 4 = \dfrac{1}{2}\ln 2\).
Answer: \(\dfrac{1}{2}\ln 2\)
Worked example Integrating tan x
Find the exact value of \(\displaystyle\int_0^{\pi/3} \tan x\,dx\).
  1. \(\tan x = \dfrac{\sin x}{\cos x} = -\dfrac{-\sin x}{\cos x}\), and \(-\sin x\) is the derivative of \(\cos x\).
  2. \(\displaystyle\int \tan x\,dx = -\ln|\cos x|\).
  3. \(-\ln\cos\dfrac{\pi}{3} + \ln\cos 0 = -\ln\dfrac{1}{2} + 0 = \ln 2\).
Answer: \(\ln 2\)
Examiner trap

\(\displaystyle\int \dfrac{x}{x^2+4}\,dx\) is not \(\ln(x^2+4)\): that differentiates to \(\dfrac{2x}{x^2+4}\). Always differentiate your log mentally and compare the numerator, then fix the constant in front.

Section 02

Linear factors: split, then integrate.

If the numerator is not a multiple of the derivative, decompose into partial fractions (the forms from section 3.1), find the constants by substituting convenient values of \(x\), then integrate each piece.

DenominatorPartial fractions
\((ax+b)(cx+d)\)\(\dfrac{A}{ax+b} + \dfrac{B}{cx+d}\)
\((ax+b)(cx+d)^2\)\(\dfrac{A}{ax+b} + \dfrac{B}{cx+d} + \dfrac{C}{(cx+d)^2}\)
\((ax+b)(cx^2+d)\)\(\dfrac{A}{ax+b} + \dfrac{Bx+C}{cx^2+d}\)
Worked example Two linear factors
Find the exact value of \(\displaystyle\int_0^1 \dfrac{5x+3}{(x+1)(2x+1)}\,dx\).
  1. Write \(\dfrac{5x+3}{(x+1)(2x+1)} = \dfrac{A}{x+1} + \dfrac{B}{2x+1}\), so \(5x + 3 = A(2x+1) + B(x+1)\).
  2. \(x = -1\): \(-2 = -A\), so \(A = 2\). \(x = -\dfrac{1}{2}\): \(\dfrac{1}{2} = \dfrac{1}{2}B\), so \(B = 1\).
  3. Integrate: \(\left[2\ln(x+1) + \dfrac{1}{2}\ln(2x+1)\right]_0^1\).
  4. \(= 2\ln 2 + \dfrac{1}{2}\ln 3 - 0\). Combine if asked: \(\ln 4 + \ln\sqrt{3} = \ln\left(4\sqrt{3}\right)\).
Answer: \(2\ln 2 + \dfrac{1}{2}\ln 3 = \ln\left(4\sqrt{3}\right)\) (about 1.94)
Your turn
Find \(\displaystyle\int \dfrac{2}{x^2-1}\,dx\).
Examiner trap

\(\dfrac{1}{2x+1}\) integrates to \(\dfrac{1}{2}\ln|2x+1|\), not \(\ln|2x+1|\). Partial fractions with a coefficient of \(x\) other than 1 are where this mark is most often lost.

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Section 03

Repeated and quadratic factors: three kinds of answer.

A squared factor gives a term that integrates by the power rule, not to a log. A quadratic factor \(x^2 + k^2\) gives a numerator \(Bx + C\) that splits into a log part and an inverse tan part.

Worked example A repeated factor
Find the exact value of \(\displaystyle\int_2^3 \dfrac{3x+1}{(x-1)(x+1)^2}\,dx\).
  1. \(3x + 1 = A(x+1)^2 + B(x-1)(x+1) + C(x-1)\).
  2. \(x = 1\): \(4 = 4A\), so \(A = 1\). \(x = -1\): \(-2 = -2C\), so \(C = 1\). Coefficients of \(x^2\): \(0 = A + B\), so \(B = -1\).
  3. Integrand \(= \dfrac{1}{x-1} - \dfrac{1}{x+1} + \dfrac{1}{(x+1)^2}\).
  4. \(\displaystyle\int \dfrac{1}{(x+1)^2}\,dx = \int (x+1)^{-2}\,dx = -\dfrac{1}{x+1}\).
  5. \(\left[\ln(x-1) - \ln(x+1) - \dfrac{1}{x+1}\right]_2^3 = \left(\ln 2 - \ln 4 - \dfrac{1}{4}\right) - \left(0 - \ln 3 - \dfrac{1}{3}\right)\).
  6. \(= \ln\dfrac{2\times 3}{4} + \dfrac{1}{3} - \dfrac{1}{4}\).
Answer: \(\ln\dfrac{3}{2} + \dfrac{1}{12}\)
Worked example A quadratic factor
Find the exact value of \(\displaystyle\int_0^1 \dfrac{3x^2+x}{(x+2)(x^2+1)}\,dx\).
  1. \(3x^2 + x = A(x^2+1) + (Bx+C)(x+2)\).
  2. \(x = -2\): \(10 = 5A\), so \(A = 2\). \(x^2\): \(3 = A + B\), so \(B = 1\). Constants: \(0 = A + 2C\), so \(C = -1\).
  3. Integrand \(= \dfrac{2}{x+2} + \dfrac{x}{x^2+1} - \dfrac{1}{x^2+1}\).
  4. Integrate: \(2\ln(x+2) + \dfrac{1}{2}\ln(x^2+1) - \tan^{-1}x\).
  5. Limits: \(\left(2\ln 3 + \dfrac{1}{2}\ln 2 - \dfrac{\pi}{4}\right) - 2\ln 2 = 2\ln 3 - \dfrac{3}{2}\ln 2 - \dfrac{\pi}{4}\).
Answer: \(2\ln 3 - \dfrac{3}{2}\ln 2 - \dfrac{\pi}{4}\) (about 0.372)
Examiner trap

\(\dfrac{1}{(x+1)^2}\) does NOT integrate to a logarithm. Write it as \((x+1)^{-2}\) and use the power rule: \(-\dfrac{1}{x+1}\). Likewise \(\dfrac{1}{x^2+1}\) gives \(\tan^{-1}x\), not \(\ln(x^2+1)\).

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Six original Cambridge-style questions
f'(x)/f(x) logarithms, tan-type integrals, linear, repeated and quadratic factors, an improper fraction and a flow-rate model.