AcademyIntegral Calculus
Academy
Integration By Parts
Level 1 - Calculus topic page in Integral Calculus.
Principle
Integration by parts reverses the product rule. It is useful when an integrand is a product and differentiating one factor simplifies it.
Notation
\(u\)
factor chosen to be differentiated
\(dv\)
factor chosen to be integrated
\(v\)
an antiderivative of dv
The Core Method
Choose \(u\) and \(dv\), then compute \(du\) and \(v\).
Integration by parts
\[\int u\,dv=uv-\int v\,du\]
A good choice makes the remaining integral simpler than the original one.
Worked Cases
Question
Find \(\int x e^x\,dx\).
Answer
Choose \(u=x\), so \(du=dx\), and choose \(dv=e^x dx\), so \(v=e^x\). Then \(\int x e^x dx=xe^x-\int e^x dx=xe^x-e^x+C\).
Examples
Question
Find \(\int \ln x\,dx\) for \(x>0\).
Answer
Write \(\ln x\) as \(\ln x\cdot1\). Choose \(u=\ln x\), so \(du=dx/x\), and \(dv=dx\), so \(v=x\). Then \(\int\ln x\,dx=x\ln x-\int1\,dx=x\ln x-x+C\).