Method guideCalculus IIIntermediate11 min

Integration by parts: choosing u and knowing when to repeat

Reverse the product rule, choose the factor that simplifies under differentiation, and recognize when the method should be repeated or abandoned.

LaTeX article Updated July 13, 2026

\int u\,dv=uv-\int v\,du

The product rule runs backward

Differentiating uv produces u dv + v du in differential notation. Rearranging and integrating gives the by-parts identity.

The method trades one integral for another. It succeeds only when the new integral is simpler or creates an equation involving the original integral.

d(uv)=u\,dv+v\,du

Choose u and dv as a pair

A good u becomes simpler when differentiated. A good dv includes the remaining factors and can be integrated without creating a worse problem.

LIATE—logarithmic, inverse trig, algebraic, trig, exponential—often suggests u, but always verify that the resulting v and remaining integral improve the structure.

\int x e^x\,dx:\quad u=x,\ dv=e^x\,dx

Repeated use should have a destination

Polynomial factors may require repeated integration by parts until differentiation reaches zero. Tabular bookkeeping can shorten that repetition.

Cyclic integrals can return the original unknown integral. Collect it algebraically instead of continuing forever, as in the classic sec³x calculation.

I=\int e^x\cos x\,dx\quad\Rightarrow\quad I=\text{known terms}-I

Worked example

Common mistakes

Choosing dv that cannot be integrated cleanly.
Forgetting the minus sign in the formula.
Stopping with a remaining integral that is not actually simpler.

Keep these ideas

By parts reverses the product rule.
Choose u and dv together.
Repeated use needs a simplifying pattern.