The shape of the lesson
By the end of the lesson, learners can
- Top is the derivative of the bottom: a logarithm: Differentiate \(\ln|f(x)|\) with the chain rule and you get \(\dfrac{f'(x)}{f(x)}\).
- 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.
- Repeated and quadratic factors: three kinds of answer: A squared factor gives a term that integrates by the power rule, not to a log.
- Check first: if the numerator is \(k f'(x)\), the answer is \(k\ln|f(x)|\).
- Improper? If the top has the same degree as the bottom, start with a constant term.
- Split into partial fractions; substitute roots, then compare coefficients.
- Integrate each piece: \(\ln\) for linear, power rule for squared, \(\ln\) plus \(\tan^{-1}\) for \(x^2 + k^2\).
- Tidy with the log laws; give exact values unless asked otherwise.
One picture every idea returns to
\(\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c\). 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.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Split it, then log it.. Paper 3 only. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Top is the derivative of the bottom: a logarithm | Differentiate \(\ln|f(x)|\) with the chain rule and you get \(\dfrac{f'(x)}{f(x)}\). Model the worked example Fix the constant on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: 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. Model the worked example Two linear factors on the board, one line at a time. Your turn: Find \(\displaystyle\int \dfrac{2}{x^2-1}\,dx\). | Pairs |
| 21 to 30 min | Teach and check: Repeated and quadratic factors: three kinds of answer | A squared factor gives a term that integrates by the power rule, not to a log. Model the worked example A repeated factor on the board, one line at a time. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 30 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Repeated and quadratic factors: three kinds of answer, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(\displaystyle\int \dfrac{x}{x^2+4}\,dx\) 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. |
| \(\dfrac{1}{2x+1}\) 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. |
| \(\dfrac{1}{(x+1)^2}\) 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 \(\ln(x^2+1)\). |
Integration using partial fractions and f'(x)/f(x): worksheet
6 questions, 30 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\displaystyle\int \dfrac{7x-1}{(x+1)(x-3)}\,dx\).
Find the exact value of \(\displaystyle\int_0^1 \dfrac{e^{x}}{e^{x}+1}\,dx\).
(a) Express \(\dfrac{2x^2+2}{(x+1)(x-1)^2}\) in partial fractions.
(b) Hence show that \(\displaystyle\int_2^3 \dfrac{2x^2+2}{(x+1)(x-1)^2}\,dx = 1 + \ln\dfrac{8}{3}\).
Show that \(\displaystyle\int_0^{\pi/3} \dfrac{\sin x}{1+\cos x}\,dx = \ln\dfrac{4}{3}\).
(a) Express \(\dfrac{x^2+2}{(x-1)(x+2)}\) in the form \(1 + \dfrac{A}{x-1} + \dfrac{B}{x+2}\).
(b) Hence find the exact value of \(\displaystyle\int_2^3 \dfrac{x^2+2}{(x-1)(x+2)}\,dx\).
Water flows into a tank at a rate of \(r\) litres per minute, where
\(r = \dfrac{4t^2+3t+9}{(2t+1)(t^2+4)}\) and \(t\) is the time in minutes.
(a) Express \(r\) in partial fractions in the form \(\dfrac{A}{2t+1} + \dfrac{Bt+C}{t^2+4}\).
(b) Find the exact volume of water that flows in during the first 2 minutes, and give it to 3 significant figures.
Answer key and marking notes
- \(7x - 1 = A(x-3) + B(x+1)\).
- \(x = 3\): \(20 = 4B\), \(B = 5\). \(x = -1\): \(-8 = -4A\), \(A = 2\).
- \(\displaystyle\int \left(\dfrac{2}{x+1} + \dfrac{5}{x-3}\right) dx = 2\ln|x+1| + 5\ln|x-3| + c\).
- The top is the derivative of the bottom, so the integral is \(\ln(e^{x}+1)\).
- \(\ln(e+1) - \ln 2 = \ln\dfrac{e+1}{2}\).
- (a) \(2x^2 + 2 = A(x-1)^2 + B(x+1)(x-1) + C(x+1)\).
- \(x = 1\): \(4 = 2C\), \(C = 2\). \(x = -1\): \(4 = 4A\), \(A = 1\). \(x^2\): \(2 = A + B\), \(B = 1\).
- \(\dfrac{1}{x+1} + \dfrac{1}{x-1} + \dfrac{2}{(x-1)^2}\).
- (b) \(\left[\ln(x+1) + \ln(x-1) - \dfrac{2}{x-1}\right]_2^3 = (\ln 4 + \ln 2 - 1) - (\ln 3 + 0 - 2)\).
- \(= \ln 8 - \ln 3 + 1 = 1 + \ln\dfrac{8}{3}\).
- The derivative of \(1 + \cos x\) is \(-\sin x\), so \(\displaystyle\int \dfrac{\sin x}{1+\cos x}\,dx = -\ln(1+\cos x)\).
- Upper: \(-\ln\left(1 + \dfrac{1}{2}\right) = -\ln\dfrac{3}{2}\). Lower: \(-\ln 2\).
- \(-\ln\dfrac{3}{2} + \ln 2 = \ln\dfrac{2}{3/2} = \ln\dfrac{4}{3}\).
- (a) \(x^2 + 2 = (x-1)(x+2) + A(x+2) + B(x-1)\).
- \(x = 1\): \(3 = 3A\), \(A = 1\). \(x = -2\): \(6 = -3B\), \(B = -2\).
- So \(1 + \dfrac{1}{x-1} - \dfrac{2}{x+2}\).
- (b) \(\left[x + \ln(x-1) - 2\ln(x+2)\right]_2^3 = (3 + \ln 2 - 2\ln 5) - (2 + 0 - 2\ln 4)\).
- \(= 1 + \ln 2 + 2\ln\dfrac{4}{5} = 1 + \ln\dfrac{32}{25}\).
- (a) \(4t^2 + 3t + 9 = A(t^2+4) + (Bt+C)(2t+1)\).
- \(t = -\dfrac{1}{2}\): \(1 - \dfrac{3}{2} + 9 = \dfrac{17}{2} = \dfrac{17}{4}A\), so \(A = 2\).
- \(t^2\): \(4 = A + 2B\), \(B = 1\). Constants: \(9 = 4A + C\), \(C = 1\). So \(r = \dfrac{2}{2t+1} + \dfrac{t+1}{t^2+4}\).
- (b) \(\displaystyle\int \dfrac{2}{2t+1}\,dt = \ln(2t+1)\); \(\displaystyle\int \dfrac{t}{t^2+4}\,dt = \dfrac{1}{2}\ln(t^2+4)\); \(\displaystyle\int \dfrac{1}{t^2+4}\,dt = \dfrac{1}{2}\tan^{-1}\dfrac{t}{2}\).
- \(\left[\ln(2t+1) + \dfrac{1}{2}\ln(t^2+4) + \dfrac{1}{2}\tan^{-1}\dfrac{t}{2}\right]_0^2 = \ln 5 + \dfrac{1}{2}\ln 8 + \dfrac{\pi}{8} - \dfrac{1}{2}\ln 4\).
- \(= \ln 5 + \dfrac{1}{2}\ln 2 + \dfrac{\pi}{8} = 2.35\) litres (3 s.f.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^1 \dfrac{4x}{x^2 + 1}\,dx\).
- A\(2\ln 2\)
- B\(4\ln 2\)
- C\(8\ln 2\)
- D\(\ln 2\)
(Paper 3 only) Find \(\displaystyle\int \dfrac{1}{(x - 1)(x + 1)}\,dx\).
- A\(\dfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right| + c\)
- B\(\dfrac{1}{2}\ln\left|\dfrac{x + 1}{x - 1}\right| + c\)
- C\(\ln|(x - 1)(x + 1)| + c\)
- D\(\ln\left|\dfrac{x - 1}{x + 1}\right| + c\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^{\frac{\pi}{6}} \dfrac{\cos x}{2 + \sin x}\,dx\).
- A\(\ln\dfrac{5}{4}\)
- B\(\ln\dfrac{5}{2}\)
- C\(-\ln\dfrac{5}{4}\)
- D\(\dfrac{1}{10}\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_1^2 \dfrac{3}{(x + 1)(x + 2)}\,dx\).
- A\(3\ln\dfrac{9}{8}\)
- B\(3\ln 2\)
- C\(3\ln\dfrac{3}{4}\)
- D\(\ln\dfrac{9}{8}\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_2^3 \dfrac{x^2 + 1}{x^2 - 1}\,dx\).
- A\(1 + \ln\dfrac{3}{2}\)
- B\(\ln\dfrac{3}{2}\)
- C\(1 + 2\ln\dfrac{3}{2}\)
- D\(1 - \ln\dfrac{3}{2}\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_0^1 \dfrac{x + 3}{(x + 1)^2}\,dx\).
- A\(1 + \ln 2\)
- B\(\ln 2 - 1\)
- C\(3\ln 2\)
- D\(\ln 2 + \dfrac{3}{2}\)
(Paper 3 only) Given that \(\dfrac{2x^2 + 2x + 2}{(x + 1)(x^2 + 1)} \equiv \dfrac{1}{x + 1} + \dfrac{x + 1}{x^2 + 1}\), find the exact value of \(\displaystyle\int_0^1 \dfrac{2x^2 + 2x + 2}{(x + 1)(x^2 + 1)}\,dx\).
- A\(\dfrac{3}{2}\ln 2 + \dfrac{\pi}{4}\)
- B\(2\ln 2 + \dfrac{\pi}{4}\)
- C\(\ln 2 + \dfrac{\pi}{4}\)
- D\(\dfrac{5}{2}\ln 2\)
(Paper 3 only) Find the exact value of \(\displaystyle\int_1^2 \dfrac{x + 5}{x(x + 1)}\,dx\), giving your answer in the form \(\ln k\).
- A\(\ln\dfrac{512}{81}\)
- B\(\ln 162\)
- C\(\ln\dfrac{32}{81}\)
- D\(\ln\dfrac{243}{512}\)
Quiz key
\(\dfrac{4x}{x^2 + 1} = 2 \times \dfrac{2x}{x^2 + 1}\), so the integral is \(\left[2\ln(x^2 + 1)\right]_0^1 = 2\ln 2 - 2\ln 1 = 2\ln 2\). The tempting \(4\ln 2\) uses the 4 as the multiplier without allowing for the 2 in \(2x\). \(8\ln 2\) multiplies by 2 instead of dividing, and \(\ln 2\) drops the constant multiple altogether.
\(A = \dfrac{1}{2}\) (put \(x = 1\)) and \(B = -\dfrac{1}{2}\) (put \(x = -1\)). So the integral is \(\dfrac{1}{2}\ln|x - 1| - \dfrac{1}{2}\ln|x + 1| + c = \dfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right| + c\). The tempting reversed fraction comes from swapping the signs of \(A\) and \(B\). \(\ln|(x - 1)(x + 1)|\) treats the integrand as \(\dfrac{f'(x)}{f(x)}\), but the top is 1, not \(2x\), and the last option forgets the \(\dfrac{1}{2}\).
\(\dfrac{d}{dx}(2 + \sin x) = \cos x\), so the integral is \(\left[\ln(2 + \sin x)\right]_0^{\pi/6} = \ln\dfrac{5}{2} - \ln 2 = \ln\dfrac{5}{4}\). The tempting \(\ln\dfrac{5}{2}\) forgets the lower limit, which is \(\ln 2\), not 0. \(-\ln\dfrac{5}{4}\) thinks the derivative of \(\sin x\) is \(-\cos x\), and \(\dfrac{1}{10}\) integrates \((2 + \sin x)^{-1}\) as if it were a power, \(-(2 + \sin x)^{-1}\).
\(\left[3\ln(x + 1) - 3\ln(x + 2)\right]_1^2 = 3\ln\dfrac{3}{4} - 3\ln\dfrac{2}{3} = 3\ln\left(\dfrac{3}{4} \times \dfrac{3}{2}\right) = 3\ln\dfrac{9}{8}\). The tempting \(3\ln 2\) uses \(+\dfrac{3}{x + 2}\), a sign slip in the partial fractions. \(3\ln\dfrac{3}{4}\) forgets the lower limit, and \(\ln\dfrac{9}{8}\) loses the 3 from the numerator.
\(\dfrac{2}{(x - 1)(x + 1)} = \dfrac{1}{x - 1} - \dfrac{1}{x + 1}\). So the integral is \(\left[x + \ln(x - 1) - \ln(x + 1)\right]_2^3 = (3 + \ln 2 - \ln 4) - (2 + \ln 1 - \ln 3) = 1 + \ln\dfrac{3}{2}\). The tempting \(\ln\dfrac{3}{2}\) forgets the constant 1 from the equal-degree split. \(1 + 2\ln\dfrac{3}{2}\) takes the partial fraction constants as \(\pm 2\) instead of \(\pm 1\), and \(1 - \ln\dfrac{3}{2}\) swaps their signs.
\(\displaystyle\int \dfrac{2}{(x + 1)^2}\,dx = -\dfrac{2}{x + 1}\). So \(\left[\ln(x + 1) - \dfrac{2}{x + 1}\right]_0^1 = (\ln 2 - 1) - (0 - 2) = 1 + \ln 2\). The tempting \(\ln 2 - 1\) integrates \(2(x + 1)^{-2}\) to \(+\dfrac{2}{x + 1}\), a sign slip. \(3\ln 2\) integrates \(\dfrac{2}{(x + 1)^2}\) as a logarithm, and \(\ln 2 + \dfrac{3}{2}\) leaves the square in the answer, \(-\dfrac{2}{(x + 1)^2}\).
\(\displaystyle\int_0^1 \dfrac{1}{x + 1}\,dx = \ln 2\); \(\displaystyle\int_0^1 \dfrac{x}{x^2 + 1}\,dx = \left[\dfrac{1}{2}\ln(x^2 + 1)\right]_0^1 = \dfrac{1}{2}\ln 2\); \(\displaystyle\int_0^1 \dfrac{1}{x^2 + 1}\,dx = \tan^{-1}1 = \dfrac{\pi}{4}\). Total \(\dfrac{3}{2}\ln 2 + \dfrac{\pi}{4}\). The tempting \(2\ln 2 + \dfrac{\pi}{4}\) forgets the \(\dfrac{1}{2}\) in \(\dfrac{x}{x^2 + 1}\). \(\ln 2 + \dfrac{\pi}{4}\) leaves out the \(\dfrac{x}{x^2 + 1}\) piece, and \(\dfrac{5}{2}\ln 2\) integrates \(\dfrac{1}{x^2 + 1}\) as \(\ln(x^2 + 1)\).
\(x = 0\): \(A = 5\); \(x = -1\): \(4 = -B\), \(B = -4\). So \(\left[5\ln x - 4\ln(x + 1)\right]_1^2 = (5\ln 2 - 4\ln 3) - (0 - 4\ln 2) = 9\ln 2 - 4\ln 3 = \ln\dfrac{512}{81}\). The tempting \(\ln 162\) uses \(B = +4\). \(\ln\dfrac{32}{81}\) forgets the lower limit (\(\ln 2\), not 0, comes from \(x + 1 = 2\)), and \(\ln\dfrac{243}{512}\) swaps \(A\) and \(B\).
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.