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How do substitution, integration by parts and partial fractions let us integrate harder functions and solve differential equations?

Integrating standard functions, integration by substitution and by parts, integrating with partial fractions, and forming and solving differential equations by separating the variables.

A CCEA A-Level Mathematics answer on integrating standard exponential, logarithmic and trigonometric functions, integration by substitution and by parts, integrating with partial fractions, and solving first-order differential equations by separating the variables.

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What this dot point is asking

CCEA wants you to integrate standard functions, integrate by substitution and by parts, use partial fractions to integrate rational functions, and form and solve first-order differential equations by separating the variables. These techniques reverse the A2 differentiation and let you solve growth, decay and accumulation problems.

The answer

Standard integrals

Integration by substitution

Integration by parts

Integrating with partial fractions and differential equations

A rational function is integrated by first splitting it into partial fractions, each of which integrates to a logarithm of the form ln⁡∣x+a∣\ln|x + a|. A first-order differential equation of the form dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y) is solved by separating the variables: gather the yy terms with dydy and the xx terms with dxdx, then integrate both sides and use an initial condition to find the constant. The constant goes on one side only (the two arbitrary constants combine into one), and a single boundary condition then fixes it, turning the general solution into the particular solution the problem asks for.

Worked example: integration by substitution

Examples in context

Example 1. Population growth. The equation dPdt=kP\frac{dP}{dt} = kP separates to 1P dP=k dt\frac{1}{P}\,dP = k\,dt, integrating to ln⁡P=kt+c\ln P = kt + c and hence P=AektP = Ae^{kt}. Separation of variables derives the exponential growth law from its defining rate.

Example 2. Area under a curved graph. Finding the area under y=xexy = xe^x requires integration by parts, the reverse of differentiating the product. The harder integration techniques widen the range of areas and totals you can compute exactly.

Try this

Q1. Find ∫e4x dx\int e^{4x}\,dx. [1 mark]

  • Cue. 14e4x+c\frac{1}{4}e^{4x} + c.

Q2. Find ∫1x dx\int \frac{1}{x}\,dx. [1 mark]

  • Cue. ln⁡∣x∣+c\ln|x| + c.

Q3. Separate the variables in dydx=xy\frac{dy}{dx} = xy. [2 marks]

  • Cue. 1y dy=x dx\frac{1}{y}\,dy = x\,dx.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20206 marksFind ∫xcos⁡x dx\int x\cos x\,dx.
Show worked answer →

Use integration by parts, ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du, choosing u=xu = x (which simplifies on differentiating) and dv=cos⁡x dxdv = \cos x\,dx.

Then du=dxdu = dx and v=sin⁡xv = \sin x.

∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x−(−cos⁡x)+c=xsin⁡x+cos⁡x+c.\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x - (-\cos x) + c = x\sin x + \cos x + c.

Markers reward the correct choice of uu and dvdv, applying the by-parts formula, integrating sin⁡x\sin x, and the constant of integration.

CCEA 20197 marksSolve the differential equation dydx=2xy\frac{dy}{dx} = \frac{2x}{y}, given that y=3y = 3 when x=0x = 0.
Show worked answer →

Separate the variables: y dy=2x dxy\,dy = 2x\,dx.

Integrate both sides: ∫y dy=∫2x dx\int y\,dy = \int 2x\,dx, so y22=x2+c\frac{y^2}{2} = x^2 + c.

Use y=3y = 3, x=0x = 0: 92=0+c\frac{9}{2} = 0 + c, so c=92c = \frac{9}{2}.

Hence y22=x2+92\frac{y^2}{2} = x^2 + \frac{9}{2}, that is y2=2x2+9y^2 = 2x^2 + 9, so y=2x2+9y = \sqrt{2x^2 + 9} (taking the positive root from the initial condition).

Markers reward separating the variables, integrating both sides, using the condition to find cc, and the explicit solution.

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