A-Level Maths Integration by Parts Exam Examples
Integration by parts reverses the product rule and is useful when an integrand contains two different types of function multiplied together. The formula is ∫u dv = uv − ∫v du. Success depends on choosing the parts thoughtfully, differentiating and integrating them correctly, and checking whether the new integral is simpler than the original.
Where the integration by parts formula comes from
The product rule says d(uv)/dx = u(dv/dx) + v(du/dx). Integrate both sides and rearrange to obtain ∫u dv = uv − ∫v du. The symbols u and dv are choices you make to split the integrand, not fixed labels attached to particular functions. Choose u so that differentiating it simplifies the expression, and choose dv so that integrating it is manageable.
Work through x times e to the x
To calculate ∫x e^x dx, choose u = x and dv = e^x dx. Then du = dx and v = e^x. Substituting gives x e^x − ∫e^x dx = x e^x − e^x + C. Differentiate x e^x − e^x: by the product rule the result is e^x + x e^x − e^x = x e^x. This check confirms both the sign and the missing terms.
Integrate x sin x
For ∫x sin x dx, choose u = x, dv = sin x dx, so du = dx and v = −cos x. Then the formula gives −x cos x − ∫(−cos x) dx = −x cos x + sin x + C. The negative sign in the antiderivative of sine is easy to forget. Checking by differentiation confirms that the two cosine terms cancel, leaving x sin x.
Integrate a logarithm
The integral ∫ln x dx appears to contain only one function, but it can be written as ∫(ln x) × 1 dx. Choose u = ln x and dv = dx, so du = (1/x)dx and v = x for x > 0. Then ∫ln x dx = x ln x − ∫1 dx = x ln x − x + C. This example shows that integration by parts can be useful even when a product is not written explicitly.
Selecting which function to differentiate
A polynomial often becomes simpler when differentiated, while e^x and sine or cosine are straightforward to integrate. Try choosing the polynomial as u in products such as x² e^x. Repeated integration by parts may then be needed. A poor choice can create an integral harder than the starting one, so pause to anticipate what du and v will look like before applying the formula. Do not treat a mnemonic as a substitute for reasoning.
Check constants and definite limits
For indefinite integrals include + C. For a definite integral, find the antiderivative and evaluate it at the specified upper and lower bounds with brackets. Make sure the minus sign in uv − ∫v du is maintained when v itself is negative. Differentiate a claimed antiderivative to verify it, especially when several stages of integration by parts were involved.
Frequently asked questions
Is integration by parts always better than substitution?
No. Choose the method that makes the integrand simpler; many composite functions suit substitution instead.
Why is there a minus sign?
It comes from rearranging the integrated product rule.
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