Integration by parts comes from the product rule. If \(u(x)\) and \(v(x)\) are differentiable, then \((uv)'=u'v+uv'\). Integrate both sides and rearrange to get \[ \int u\,dv = uv - \int v\,du. \] You choose a factor to call \(u\) (differentiate it) and a factor to call \(dv\) (integrate it). The goal is a new integral that is easier than the original.

How to choose \(u\) and \(dv\)

A common classroom mnemonic is LIATE (rough priority for \(u\)): Logarithmic, Inverse trig, Algebraic, Trig, Exponential. Prefer a \(u\) that simplifies when differentiated and a \(dv\) that is easy to integrate. The mnemonic is a hint, not a theorem — if the new integral looks worse, swap choices.

Pattern 1: polynomial times exponential

For \(\int x e^{x}\,dx\), take \(u=x\), \(dv=e^{x}\,dx\). Then \(du=dx\), \(v=e^{x}\), so \[ \int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C. \] One parts step clears the polynomial factor.

Pattern 2: log without a partner

For \(\int \ln x\,dx\), write \(u=\ln x\), \(dv=dx\). Then \(du=\frac{1}{x}dx\), \(v=x\), and \[ \int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + C. \] Parts turns a lonely log into an algebraic integral.

When to stop or loop

Sometimes parts produces a similar integral (classic: \(\int e^{x}\sin x\,dx\)). Apply parts twice and solve algebraically for the original integral. Related techniques: partial fractions, separable DEs, and calculators for checks — not substitutes for understanding.

Sources

Frequently asked questions

What is integration by parts?

It is the integral form of the product rule: ∫ u dv = uv − ∫ v du. You trade one integral for another that should be easier.

What does LIATE mean?

A mnemonic for choosing u: Logarithmic, Inverse trig, Algebraic, Trig, Exponential. Use it as a starting guess, then verify the new integral is simpler.

Can I use integration by parts twice?

Yes. Some integrands return a multiple of the original integral after two applications; you then solve for that integral algebraically.

Is integration by parts the same as the product rule?

It is the product rule integrated and rearranged. Differentiating uv recovers the product rule; integrating recovers parts.