Integration by parts
Master integration by parts for Year 12 Specialist Mathematics in Queensland (QCAA). This is the technique that reverses the product rule: when an integrand is a product such as \(x e^{x}\), \(x\sin x\) or \(x\ln x\), the formula \(\int u\,dv = uv-\int v\,du\) trades it for an easier integral.
You will learn to choose \(u\) and \(dv\) with the LIATE guide, apply the formula step by step, and handle integrals that need parts more than once or that recur — a key integration technique for areas, volumes and differential equations later in the course.
Theory
Integration by parts reverses the product rule and is a core technique of Year 12 Specialist Mathematics (QCAA, Queensland). When an integrand is a product — such as \(x e^{x}\), \(x\sin x\) or \(x\ln x\) — the formula \(\int u\,dv = uv-\int v\,du\) trades it for an easier integral. This page shows how to choose \(u\) and \(dv\) with the LIATE guide, apply the formula, and handle repeated and recurring cases.
Integration by parts is the reverse of the product rule for differentiation. It is used to integrate a product of two functions where ordinary rules and substitution do not apply, for example \(x e^{x}\), \(x\sin x\), \(\ln x\) or \(\arctan x\).
You split the integrand into two parts: a factor \(u\) that you will differentiate, and a factor \(dv\) that you will integrate. From these, \(du=u'\,dx\) and \(v=\int dv\). The formula \(\int u\,dv = uv-\int v\,du\) then swaps the original integral for the usually simpler \(\int v\,du\).
The choice of \(u\) is guided by LIATE — Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Whichever type comes earlier in this list is chosen as \(u\), because differentiating it tends to simplify the problem; the other factor becomes \(dv\).
Some integrals need parts applied more than once (for example \(x^{2}e^{x}\)), and a few — such as \(e^{x}\cos x\) — return to the original integral, so you solve for it algebraically. For a definite integral, apply the same formula and evaluate \(\big[uv\big]_a^b\) at the limits.
Integration by parts, from the product rule \((uv)'=u'v+uv'\) rearranged and integrated:
Written with derivatives, exactly as the QCAA syllabus states it:
For a definite integral the boundary term is evaluated at the limits:
How to integrate by parts
- Choose \(u\) and \(dv\): use LIATE to pick \(u\) (the factor that gets simpler when differentiated); everything else, including \(dx\), is \(dv\).
- Differentiate and integrate: find \(du=u'\,dx\) and \(v=\int dv\) (no \(+c\) needed for \(v\) here).
- Apply the formula: substitute into \(\int u\,dv = uv-\int v\,du\).
- Finish the new integral: evaluate \(\int v\,du\) — if it is still a product, apply parts again; if the original integral reappears, solve for it. Add \(+c\), or evaluate the limits for a definite integral.
LIATE: \(x\) is Algebraic and \(e^{3x}\) is Exponential, so \(u=x\), \(dv=e^{3x}\,dx\):
| \(u\) | \(=\) | \(x\) |
| \(dv\) | \(=\) | \(e^{3x}\,dx\) |
| \(du\) | \(=\) | \(dx\) |
| \(v\) | \(=\) | \(\dfrac{1}{3}e^{3x}\) |
Apply \(\int u\,dv = uv-\int v\,du\), then integrate what remains:
| \(\int x e^{3x}\,dx\) | \(=\) | \(\dfrac{1}{3}x e^{3x}-\int \dfrac{1}{3}e^{3x}\,dx\) |
| \(=\) | \(\dfrac{1}{3}x e^{3x}-\dfrac{1}{9}e^{3x}+c\) |
\(\displaystyle\int x e^{3x}\,dx=\dfrac{1}{9}e^{3x}(3x-1)+c\).
LIATE picks the Algebraic factor as \(u\); the trig factor is \(dv\):
| \(u\) | \(=\) | \(x\) |
| \(dv\) | \(=\) | \(\cos 2x\,dx\) |
| \(du\) | \(=\) | \(dx\) |
| \(v\) | \(=\) | \(\dfrac{1}{2}\sin 2x\) |
Apply the by-parts formula, then integrate the remaining term:
| \(\int x\cos 2x\,dx\) | \(=\) | \(\dfrac{1}{2}x\sin 2x-\int \dfrac{1}{2}\sin 2x\,dx\) |
| \(=\) | \(\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+c\) |
\(\displaystyle\int x\cos 2x\,dx=\dfrac{1}{2}x\sin 2x+\dfrac{1}{4}\cos 2x+c\).
LIATE puts Logarithmic before Algebraic, so \(u=\ln x\), \(dv=x^{3}\,dx\):
| \(u\) | \(=\) | \(\ln x\) |
| \(dv\) | \(=\) | \(x^{3}\,dx\) |
| \(du\) | \(=\) | \(\dfrac{1}{x}\,dx\) |
| \(v\) | \(=\) | \(\dfrac{x^{4}}{4}\) |
Apply the formula; the \(\tfrac{1}{x}\) cancels a power in the new integral:
| \(\int x^{3}\ln x\,dx\) | \(=\) | \(\dfrac{x^{4}}{4}\ln x-\int \dfrac{x^{4}}{4}\cdot\dfrac{1}{x}\,dx\) |
| \(=\) | \(\dfrac{x^{4}}{4}\ln x-\int \dfrac{x^{3}}{4}\,dx\) | |
| \(=\) | \(\dfrac{x^{4}}{4}\ln x-\dfrac{x^{4}}{16}+c\) |
\(\displaystyle\int x^{3}\ln x\,dx=\dfrac{x^{4}}{4}\ln x-\dfrac{x^{4}}{16}+c\).
Take \(u=x^{2}\), \(dv=\cos x\,dx\); one application leaves a product still:
| \(u\) | \(=\) | \(x^{2}\) |
| \(dv\) | \(=\) | \(\cos x\,dx\) |
| \(du\) | \(=\) | \(2x\,dx\) |
| \(v\) | \(=\) | \(\sin x\) |
Apply parts; then reuse \(\int x\sin x\,dx=-x\cos x+\sin x\):
| \(\int x^{2}\cos x\,dx\) | \(=\) | \(x^{2}\sin x-\int 2x\sin x\,dx\) |
| \(=\) | \(x^{2}\sin x-2(-x\cos x+\sin x)\) | |
| \(=\) | \(x^{2}\sin x+2x\cos x-2\sin x+c\) |
\(\displaystyle\int x^{2}\cos x\,dx=x^{2}\sin x+2x\cos x-2\sin x+c\).
Common pitfalls
Frequently asked questions
What is integration by parts?
It is the integration rule that reverses the product rule: \(\int u\,dv = uv-\int v\,du\). It lets you integrate a product by differentiating one factor (\(u\)) and integrating the other (\(dv\)).
How do I choose \(u\) and \(dv\)?
Use the LIATE guide: pick \(u\) as the factor whose type comes first in Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. The remaining factor, together with \(dx\), is \(dv\). A good choice makes \(\int v\,du\) simpler.
Why does \(\int \ln x\,dx\) need integration by parts?
There is no direct rule for \(\int\ln x\,dx\), so write it as a product with \(dv=dx\). Then \(u=\ln x\), \(v=x\), and parts gives \(x\ln x-\int 1\,dx = x\ln x-x+c\).
When do you apply integration by parts more than once?
When the new integral \(\int v\,du\) is still a product, such as after the first step of \(\int x^{2}e^{x}\,dx\). Apply parts again to reduce the power until an ordinary integral remains.
What is the recurring or \(\text{loop}\) case?
For integrals like \(\int e^{x}\cos x\,dx\), two applications of parts bring back the original integral \(I\). Instead of looping, collect the \(I\) terms and solve algebraically, giving \(I=\tfrac{1}{2}e^{x}(\sin x+\cos x)+c\).
How do you do a definite integral by parts?
Use \(\int_a^b u\,dv=\big[uv\big]_a^b-\int_a^b v\,du\): evaluate the \(uv\) boundary term at the limits, then subtract the remaining integral evaluated over the same limits.