Short answer: use integration by parts when the integrand is a product of two different kinds of functions (for example x·ex or x·sin x) and substitution doesn't work. The formula is ∫u dv = uv − ∫v du. Choose u so that it gets simpler when you differentiate it.
The 10-second test
- Try substitution first. If part of the integrand is the derivative of another part (like 2x·cos(x²)), substitution is faster.
- Is it a product of unlike functions? A polynomial times an exponential, a trig function or a logarithm usually means parts.
- Is it a lone ln x or inverse trig function? Use parts with dv = dx. This is how you integrate ln x.
How to pick u: the LIATE rule
Pick u as the function that comes first in this list:
- Logarithmic: ln x
- Inverse trig: arctan x, arcsin x
- Algebraic: x, x², polynomials
- Trigonometric: sin x, cos x
- Exponential: ex
LIATE is a rule of thumb, not a law, but it works for most exam questions.
Worked example: ∫x·ex dx
u = x (algebraic beats exponential), so du = dx. dv = ex dx, so v = ex.
∫x·ex dx = x·ex − ∫ex dx = x·ex − ex + C.
Check it by differentiating: (x·ex − ex)′ = ex + x·ex − ex = x·ex. ✓
Worked example: ∫ln x dx
u = ln x, dv = dx, so du = (1/x) dx and v = x.
∫ln x dx = x ln x − ∫x·(1/x) dx = x ln x − x + C.
Special cases to recognise
- Repeated parts: for x²·ex, apply parts twice. Each round lowers the power of x by one. A tabular ("DI") layout keeps the signs straight.
- The loop: for ex·sin x, apply parts twice. The original integral reappears, so move it to the left side and solve for it.
Mistakes that cost marks
- Picking u = ex in x·ex: the new integral gets harder, not easier.
- Dropping the minus sign in uv − ∫v du, especially on the second round.
- Forgetting + C, or forgetting to evaluate uv at both limits in a definite integral.
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