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Calculus Integration

Integration by parts

20 practice questions 0 video lessons Theory + worked examples

Integration by Parts

California Calculus • Standard 17.0 • Integration

Integration by Parts is a topic in Integration in the California Calculus Standards. It is aligned to Standard 17.0, which requires students to compute integrals using techniques of integration such as integration by parts.

Integration by parts integrates a product using \(\int u\,dv=uv-\int v\,du\), the reverse of the product rule, with the LIATE guide for choosing \(u\).

California Calculus › Integration › Integration by Parts  —  Standard 17.0

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Theory

Integration by parts integrates a product by trading it for an easier integral: \(\int u\,dv=uv-\int v\,du\). It is the reverse of the product rule, and the LIATE guide tells you which factor to call \(u\).

Integrating the product rule \((uv)'=u'v+uv'\) and rearranging gives the integration by parts formula:

\[\int u\,dv=uv-\int v\,du.\]

You split the integrand into a part to differentiate (\(u\)) and a part to integrate (\(dv\)). A good choice makes the new integral \(\int v\,du\) simpler than the original.

The LIATE order — Logarithmic, Inverse-trig, Algebraic, Trig, Exponential — ranks which factor to choose as \(u\) (earlier in the list wins).

Key idea: pick \(u\) to be the factor that gets simpler when differentiated (like \(x\) or \(\ln x\)); the rest is \(dv\).
The integration by parts formula Integration by parts comes from the product rule and trades one integral for an easier one. ∫ u dv = uv − ∫ v du choose u by LIATE (log, inverse-trig, algebraic, trig, exp)
\(\int u\,dv=uv-\int v\,du\), with \(u\) chosen by LIATE.
The area under x times e to the x found by parts Integration by parts evaluates areas under products such as x times e to the x that no single rule handles. x y area
It finds areas under products like \(x e^{x}\).

The formula and the choice guide:

\[\int u\,dv=uv-\int v\,du\]
integral of u dv equals uv minus integral of v du
LIATE for \(u\): Logarithmic, Inverse-trig, Algebraic, Trig, Exponential — whichever appears first becomes \(u\).

How to integrate by parts

  1. Choose \(u\) and \(dv\) (use LIATE for \(u\)); the rest is \(dv\).
  2. Compute \(du\) and \(v\) by differentiating \(u\) and integrating \(dv\).
  3. Apply \(uv-\int v\,du\) and evaluate the new, simpler integral.
Example 1 — \(x e^x\)
Find \(\displaystyle\int x\,e^{x}\,dx\).
Solution

Let \(u=x\), \(dv=e^{x}\,dx\); then \(du=dx\), \(v=e^{x}\). Apply \(uv-\int v\,du\).

\(\int x e^{x}\,dx\)\(=\)\(xe^{x}-\int e^{x}\,dx\)
\(=\)\(xe^{x}-e^{x}+C\)
\(=\)\((x-1)e^{x}+C\)
integral equals x minus one times e to the x plus C
Example 2 — \(x \cos x\)
Find \(\displaystyle\int x\cos x\,dx\).
Solution

Let \(u=x\), \(dv=\cos x\,dx\); then \(du=dx\), \(v=\sin x\). Apply \(uv-\int v\,du\), using \(\int\sin x\,dx=-\cos x\).

\(\int x\cos x\,dx\)\(=\)\(x\sin x-\int \sin x\,dx\)
\(=\)\(x\sin x-(-\cos x)+C\)
\(=\)\(x\sin x+\cos x+C\)
integral equals x sin x plus cos x plus C
Example 3 — \(\ln x\)
Find \(\displaystyle\int \ln x\,dx\).
Solution

Let \(u=\ln x\), \(dv=dx\); then \(du=\dfrac{1}{x}\,dx\), \(v=x\).

\(\int \ln x\,dx\)\(=\)\(x\ln x-\int 1\,dx\)
\(=\)\(x\ln x-x+C\)
integral equals x ln x minus x plus C
Example 4 — Choose u by LIATE
In \(\displaystyle\int x\ln x\,dx\), which factor is \(u\)?
Solution

LIATE ranks logarithmic before algebraic, so \(u=\ln x\).

\(u\)\(=\)\(\ln x,\quad dv=x\,dx\)
choose u equals ln x by LIATE

Common pitfalls

Choose \(u\) wisely. A bad choice makes \(\int v\,du\) harder; LIATE keeps you on track.
Keep the minus sign. The formula subtracts \(\int v\,du\); dropping the sign is a common slip.
Some integrals need two rounds. \(\int x^2 e^{x}\,dx\) takes parts twice — keep going until the leftover integral is elementary.

Frequently asked questions

What is integration by parts?

A method to integrate a product: \(\int u\,dv=uv-\int v\,du\). It reverses the product rule and trades a hard integral for an easier one.

How do you choose u and dv?

Use LIATE: Logarithmic, Inverse-trig, Algebraic, Trig, Exponential. The factor earliest in this list is \(u\); the rest is \(dv\).

How do you integrate ln x?

By parts with \(u=\ln x\) and \(dv=dx\): the result is \(x\ln x-x+C\).

When do you use integration by parts twice?

When one application still leaves a product, such as \(\int x^2 e^{x}\,dx\). Apply parts again until the remaining integral is elementary.