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Cambridge IGCSE · Topic 14.12
IGCSE Add Maths 0606 / Topic 14 / IGCSE

Differentiation, run backwards.

Integration asks the reverse question: which function has this derivative? Add one to the power and divide by the new power, divide by \(a\) whenever the inside is \(ax + b\), and never forget the \(+\,c\), because a whole family of curves shares the same gradient function.

Topic 14All calculus lessons →
The key idea

If \(\dfrac{\mathrm{d}}{\mathrm{d}x}F(x) = f(x)\), then \(\displaystyle\int f(x)\,\mathrm{d}x = F(x) + c\). For powers, \(\displaystyle\int x^{n}\,\mathrm{d}x = \dfrac{x^{n+1}}{n + 1} + c\) (\(n \ne -1\)) and \(\displaystyle\int \dfrac{1}{x}\,\mathrm{d}x = \ln|x| + c\). When the inside is \(ax + b\), integrate as if it were \(x\), then divide by \(a\). Rewrite products, quotients and roots as sums of powers BEFORE you integrate, and check any answer by differentiating it.

x y 1 3 (1, 3) y = x² + 2 c = 2: through (1, 3) y = x² c = 0 y = x² − 2 c = −2 Every curve has dy/dx = 2x: the dashed tangents at x = 1 are all parallel (gradient 2)
Fig. 1: three members of the family y = x² + c, all with dy/dx = 2x. At x = 1 every curve has gradient 2, so the dashed tangents are parallel: the gradient function alone cannot tell the curves apart. One extra point, here (1, 3), fixes c = 2
Section 01

Differentiation in reverse.

Differentiating \(x^{3}\) gives \(3x^{2}\), so integrating \(3x^{2}\) gives back \(x^{3}\). But \(x^{3} + 5\) and \(x^{3} - 2\) also differentiate to \(3x^{2}\), because the derivative of a constant is zero. So every indefinite integral ends with an arbitrary constant: \(\displaystyle\int 3x^{2}\,\mathrm{d}x = x^{3} + c\). To reverse 'multiply by the power, then take one off the power', do the opposite steps in the opposite order: ADD one to the power, then DIVIDE by the new power.

\[ \int x^{n}\,\mathrm{d}x = \frac{x^{n+1}}{n + 1} + c \quad (n \ne -1), \qquad \int \frac{1}{x}\,\mathrm{d}x = \ln|x| + c \]

Worked example Powers, roots and reciprocals
Find (a) \(\displaystyle\int (6x^{2} - 4x + 5)\,\mathrm{d}x\) (b) \(\displaystyle\int \left(3\sqrt{x} - \frac{2}{x^{3}}\right)\mathrm{d}x\).
  1. (a) Term by term: \(\dfrac{6x^{3}}{3} - \dfrac{4x^{2}}{2} + 5x + c = 2x^{3} - 2x^{2} + 5x + c\).
  2. (b) Rewrite as powers: \(3x^{\frac{1}{2}} - 2x^{-3}\).
  3. \(3x^{\frac{1}{2}}\) becomes \(\dfrac{3x^{\frac{3}{2}}}{\frac{3}{2}} = 2x^{\frac{3}{2}}\); \(-2x^{-3}\) becomes \(\dfrac{-2x^{-2}}{-2} = x^{-2}\).
  4. Check (b): \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(2x^{\frac{3}{2}} + x^{-2}\right) = 3x^{\frac{1}{2}} - 2x^{-3}\).
Answer: (a) \(2x^{3} - 2x^{2} + 5x + c\)   (b) \(2x^{\frac{3}{2}} + \dfrac{1}{x^{2}} + c\)
Worked example Expand and split before you integrate
Find \(\displaystyle\int \frac{(x + 2)^{2}}{x}\,\mathrm{d}x\).
  1. There is no rule for a quotient, so make it a sum. Expand: \((x + 2)^{2} = x^{2} + 4x + 4\).
  2. Split over the single term \(x\): \(\dfrac{x^{2}}{x} + \dfrac{4x}{x} + \dfrac{4}{x} = x + 4 + \dfrac{4}{x}\).
  3. Integrate: \(\dfrac{x^{2}}{2} + 4x + 4\ln|x| + c\).
  4. Check: \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{x^{2}}{2} + 4x + 4\ln|x|\right) = x + 4 + \dfrac{4}{x}\).
Answer: \(\dfrac{x^{2}}{2} + 4x + 4\ln|x| + c\)
Your turn
Find \(\displaystyle\int \left(4x^{3} - \frac{1}{x^{2}}\right)\mathrm{d}x\).
Examiner trap

You cannot integrate the top and bottom of a fraction separately, or the factors of a product separately: \(\displaystyle\int \frac{(x + 2)^{2}}{x}\,\mathrm{d}x\) is NOT \(\dfrac{(x + 2)^{3}/3}{x^{2}/2}\). Expand and split first. And only \(x^{-1}\) gives a logarithm: \(\displaystyle\int \frac{1}{x^{2}}\,\mathrm{d}x = -\frac{1}{x} + c\), not \(\ln x^{2}\).

Section 02

A linear inside: divide by a.

When you differentiate a function of \(ax + b\), the chain rule MULTIPLIES by \(a\): \(\dfrac{\mathrm{d}}{\mathrm{d}x}(3x - 2)^{6} = 18(3x - 2)^{5}\). Integration undoes this, so integrate as though \(ax + b\) were just \(x\), then DIVIDE by \(a\). This only works because \(a\) is a constant; it does not work for an inside such as \(x^{2} + 1\). Angles are in radians.

\(f(x)\)\(\displaystyle\int f(x)\,\mathrm{d}x\)Check: the derivative you reversed
\((ax + b)^{n}\), \(n \ne -1\)\(\dfrac{(ax + b)^{n+1}}{a(n + 1)} + c\)\(\dfrac{\mathrm{d}}{\mathrm{d}x}(ax + b)^{n+1} = a(n + 1)(ax + b)^{n}\)
\(\dfrac{1}{ax + b}\)\(\dfrac{1}{a}\ln|ax + b| + c\)\(\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(ax + b) = \dfrac{a}{ax + b}\)
\(\mathrm{e}^{ax + b}\)\(\dfrac{1}{a}\mathrm{e}^{ax + b} + c\)\(\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{ax + b} = a\,\mathrm{e}^{ax + b}\)
\(\sin(ax + b)\)\(-\dfrac{1}{a}\cos(ax + b) + c\)\(\dfrac{\mathrm{d}}{\mathrm{d}x}\cos(ax + b) = -a\sin(ax + b)\)
\(\cos(ax + b)\)\(\dfrac{1}{a}\sin(ax + b) + c\)\(\dfrac{\mathrm{d}}{\mathrm{d}x}\sin(ax + b) = a\cos(ax + b)\)
\(\sec^{2}(ax + b)\)\(\dfrac{1}{a}\tan(ax + b) + c\)\(\dfrac{\mathrm{d}}{\mathrm{d}x}\tan(ax + b) = a\sec^{2}(ax + b)\)
Worked example Powers of a bracket
Find (a) \(\displaystyle\int (3x - 2)^{5}\,\mathrm{d}x\) (b) \(\displaystyle\int \frac{4}{\sqrt{2x + 1}}\,\mathrm{d}x\).
  1. (a) Add one to the power and divide by the new power AND by \(a = 3\): \(\dfrac{(3x - 2)^{6}}{3 \times 6} = \dfrac{(3x - 2)^{6}}{18}\).
  2. (b) Rewrite: \(4(2x + 1)^{-\frac{1}{2}}\). The new power is \(\frac{1}{2}\) and \(a = 2\).
  3. \(\dfrac{4(2x + 1)^{\frac{1}{2}}}{2 \times \frac{1}{2}} = 4(2x + 1)^{\frac{1}{2}}\).
  4. Check (b): \(\dfrac{\mathrm{d}}{\mathrm{d}x}4(2x + 1)^{\frac{1}{2}} = 4 \times \dfrac{1}{2}(2x + 1)^{-\frac{1}{2}} \times 2 = \dfrac{4}{\sqrt{2x + 1}}\).
Answer: (a) \(\dfrac{(3x - 2)^{6}}{18} + c\)   (b) \(4\sqrt{2x + 1} + c\)
Worked example Exponential, logarithm and trigonometry
Find (a) \(\displaystyle\int \left(\mathrm{e}^{4x - 1} + \frac{6}{3x + 1}\right)\mathrm{d}x\) (b) \(\displaystyle\int \left(\sin 3x + 2\sec^{2} 2x\right)\mathrm{d}x\).
  1. (a) \(\mathrm{e}^{4x - 1}\) integrates to \(\dfrac{1}{4}\mathrm{e}^{4x - 1}\).
  2. \(\dfrac{6}{3x + 1}\) integrates to \(6 \times \dfrac{1}{3}\ln|3x + 1| = 2\ln|3x + 1|\).
  3. (b) \(\sin 3x\) integrates to \(-\dfrac{1}{3}\cos 3x\): the minus sign comes from \(\dfrac{\mathrm{d}}{\mathrm{d}x}\cos 3x = -3\sin 3x\).
  4. \(2\sec^{2} 2x\) integrates to \(2 \times \dfrac{1}{2}\tan 2x = \tan 2x\).
Answer: (a) \(\dfrac{1}{4}\mathrm{e}^{4x - 1} + 2\ln|3x + 1| + c\)   (b) \(-\dfrac{1}{3}\cos 3x + \tan 2x + c\)
Your turn
Find \(\displaystyle\int 8\cos(4x + 1)\,\mathrm{d}x\).
Examiner trap

Divide by \(a\), never multiply: \(\displaystyle\int \mathrm{e}^{3x}\,\mathrm{d}x = \frac{1}{3}\mathrm{e}^{3x} + c\), not \(3\mathrm{e}^{3x}\) (that is the derivative). Watch the signs too: \(\sin\) integrates to \(-\cos\), and a negative \(a\) flips the sign, as \(\displaystyle\int \mathrm{e}^{-2x}\,\mathrm{d}x = -\frac{1}{2}\mathrm{e}^{-2x} + c\).

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Find \(\displaystyle\int x^{4}\,\mathrm{d}x\).

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Find \(\displaystyle\int \mathrm{e}^{3x}\,\mathrm{d}x\).

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Find \(\displaystyle\int \cos 2x\,\mathrm{d}x\).

Section 03

Pinning down c, and 'hence' questions.

Integrating \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) gives a whole family of curves (Fig. 1). If you also know one point on the curve, substitute it into \(y = F(x) + c\) to find \(c\) and the equation of THE curve. Exam questions also give you a derivative to work out first, then say 'hence' integrate: rearrange the derivative so that the function you want is integrated by reading the result backwards.

Worked example Equation of a curve from its gradient
A curve has \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - \dfrac{4}{x^{2}}\) and passes through the point \((2, 5)\). Find the equation of the curve.
  1. Write \(3x^{2} - 4x^{-2}\) and integrate: \(y = x^{3} - \dfrac{4x^{-1}}{-1} + c = x^{3} + \dfrac{4}{x} + c\).
  2. Substitute \(x = 2\), \(y = 5\): \(5 = 8 + 2 + c\).
  3. So \(c = -5\).
  4. Check: at \(x = 2\), \(y = 8 + 2 - 5 = 5\), as required.
Answer: \(y = x^{3} + \dfrac{4}{x} - 5\)
Worked example Differentiate, then hence integrate
Find \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x\ln x)\). Hence find \(\displaystyle\int \ln x\,\mathrm{d}x\).
  1. Product rule: \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x\ln x) = 1 \times \ln x + x \times \dfrac{1}{x} = \ln x + 1\).
  2. Read it backwards: \(\displaystyle\int (\ln x + 1)\,\mathrm{d}x = x\ln x + c\).
  3. The extra term \(1\) integrates to \(x\), so \(\displaystyle\int \ln x\,\mathrm{d}x = x\ln x - \int 1\,\mathrm{d}x = x\ln x - x + c\).
  4. Check: \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x\ln x - x) = \ln x + 1 - 1 = \ln x\).
Answer: \(\ln x + 1\); \(\displaystyle\int \ln x\,\mathrm{d}x = x\ln x - x + c\)
Your turn
Show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x\sin x) = \sin x + x\cos x\), and hence find \(\displaystyle\int x\cos x\,\mathrm{d}x\).
Examiner trap

Find \(c\) AFTER integrating, by substituting into \(y\), never into \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\); and the \(y\)-coordinate of the point is not \(c\) unless \(x = 0\). In a 'hence' question, deal with the extra terms: \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x\ln x) = \ln x + 1\) does NOT mean \(\displaystyle\int \ln x\,\mathrm{d}x = x\ln x\).

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Find \(\displaystyle\int \frac{x^{3} + 4}{x^{2}}\,\mathrm{d}x\).

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Find \(\displaystyle\int \frac{6}{2x - 5}\,\mathrm{d}x\).

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A curve has \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x - 3\) and passes through \((2, 7)\). Find the equation of the curve.

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Given that \(\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x^{2} + 1) = \dfrac{2x}{x^{2} + 1}\), find \(\displaystyle\int \frac{x}{x^{2} + 1}\,\mathrm{d}x\).

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Derivatives reversed, constants found.

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Go deeper · practice
Six original Cambridge-style questions
Powers and 1/x, rewriting a quotient of roots, linear insides with negative a, a mixed trigonometric integral, the equation of a curve through a point, and a 'hence' integral of x ln x.