Integration by Parts

Integration by Parts

IB Year 6 | Grade 12
TGM Original

Reverse the product rule when a product does not simplify by inspection or substitution.

\(\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u\)

Choose \(u\) to become simpler when differentiated, and \(\mathrm{d}v\) to have an accessible antiderivative. LIATE (logarithmic, inverse trigonometric, algebraic, trigonometric, exponential) is a useful guide, not a compulsory rule.

\(\begin{aligned}u&=\ln x,\qquad\mathrm{d}v=x\,\mathrm{d}x\\\mathrm{d}u&=\dfrac{1}{x}\,\mathrm{d}x,\qquad v=\dfrac{x^2}{2}\end{aligned}\)

\(\begin{aligned}\int x\ln x\,\mathrm{d}x&=\dfrac{x^2}{2}\ln x-\dfrac{1}{2}\int x\,\mathrm{d}x\\&=\dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}+C\qquad(x>0)\end{aligned}\)

AHL5.16 includes repeated integration by parts. A polynomial factor can require several rounds; an exponential–trigonometric product can return to the original integral, which you solve for algebraically.

Definite integration by parts

\(\int_a^b\mathrm{u}(x)\mathrm{v}^{\prime}(x)\,\mathrm{d}x=[\mathrm{u}(x)\mathrm{v}(x)]_a^b-\int_a^b\mathrm{v}(x)\mathrm{u}^{\prime}(x)\,\mathrm{d}x\)

Apply the limits to the product term as well as to the remaining integral. No arbitrary constant remains in a definite value.

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Need help? Join our JC Math tuition classes.

Learn more