Integration by Parts

Level 1 - Math I (Physics) topic page in Integration.

Integration by parts integrates products of functions. It's based on the product rule for differentiation reversed.

IBP Formula

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

This is derived from:

Derivation

\[\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \Rightarrow u\,dv = d(uv) - v\,du\]

Use the LIATE rule (prioritize in this order):

Example 1: \(\int x e^x \, dx\)

Let \(u = x\) (algebraic), \(dv = e^x \, dx\) (exponential):

Example 1

\[u = x \Rightarrow du = dx, \quad dv = e^x \, dx \Rightarrow v = e^x\]

Result 1

\[\int x e^x \, dx = xe^x - \int e^x \, dx = xe^x - e^x + C = e^x(x - 1) + C\]

Example 2: \(\int \ln x \, dx\)

Let \(u = \ln x\) (logarithmic), \(dv = dx\):

Example 2

\[u = \ln x \Rightarrow du = \frac{1}{x}dx, \quad v = x\]

Result 2

\[\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x}dx = x\ln x - \int 1 \, dx = x\ln x - x + C\]

For integrals like \(\int e^x \sin x \, dx\), apply IBP twice:

Repeated

\[\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx\]

Then apply IBP again to the remaining integral and solve for the original integral algebraically.

For repeated differentiation of one factor, use a table:

This efficiently handles products like \(\int x^2 e^x \, dx\).