Integration by Parts

You’ll be able to

Apply the integration-by-parts formula to products of functions
Choose u and dv strategically using the LIATE guideline
Handle repeated integration by parts and cyclic cases

The product rule, run backward

Integration by parts is the integration counterpart of the product rule. It rewrites a hard integral ∫u dv as u·v − ∫v du, trading one integral for another that is (ideally) easier. It is the go-to method for products of unlike functions — a polynomial times an exponential, a polynomial times a trig function, or a lone logarithm. The whole game is choosing u and dv so the new integral ∫v du is simpler than the one you started with.

Integration by parts

∫ u dv = u·v − ∫ v du

You pick u (to differentiate) and dv (to integrate); then du = u′ dx and v = ∫dv.

Choosing u with LIATE

Pick u to be whatever comes first in LIATE: Logarithmic, Inverse trig, Algebraic (polynomial), Trigonometric, Exponential. Whatever is left becomes dv. The logic: you want u to become simpler when differentiated (a polynomial eventually differentiates to a constant) and dv to be something you can readily integrate. For ∫x·eˣ dx, "algebraic" x beats "exponential" eˣ, so u = x and dv = eˣ dx.

Worked example

Evaluate ∫ x·eˣ dx.

1.By LIATE, algebraic beats exponential, so let u = x and dv = eˣ dx.
2.Then du = dx and v = ∫eˣ dx = eˣ.
3.Apply the formula: ∫x·eˣ dx = u·v − ∫v du = x·eˣ − ∫eˣ dx.
4.Integrate the remaining piece: x·eˣ − eˣ + C.

Answer: ∫ x·eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C. The new integral ∫eˣ dx was easier than the original.

Tip

To integrate a lone logarithm like ∫ln(x) dx, set u = ln(x) and dv = dx. Then du = (1/x) dx and v = x, giving x·ln(x) − ∫1 dx = x·ln(x) − x + C. The trick is realizing dv can just be dx.

Checkpoint

For ∫ x·cos(x) dx, which choice of u and dv follows the LIATE guideline?

Checkpoint

Using integration by parts with u = x and dv = eˣ dx, what is ∫ x·eˣ dx?

On the exam

Some integrals like ∫eˣ·sin(x) dx cycle back to the original after two rounds of parts. When that happens, set the integral equal to I, solve the algebraic equation for I, and divide — do not keep integrating forever.

Answer the 2 checkpoints as you read.