Integration by Parts Examples in Math
Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Integration by Parts.
This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.
Concept Recap
An integration technique based on the product rule: \int u\,dv = uv - \int v\,du. Used when the integrand is a product of two functions.
The product rule for derivatives says (uv)' = u'v + uv'. Rearranging and integrating gives integration by parts. The idea is to trade your original integral for a (hopefully easier) one. You're transferring the derivative from one factor to the other.
Read the full concept explanation โHow to Use These Examples
- Read the first worked example with the solution open so the structure is clear.
- Try the practice problems before revealing each solution.
- Use the related concepts and background knowledge badges if you feel stuck.
What to Focus On
Core idea: Integration by parts converts \int u\,dv into uv - \int v\,du. The goal is to choose u and dv so that \int v\,du is simpler than the original integral.
Common stuck point: Sometimes you need to apply integration by parts TWICE (as with \int e^x\sin x\,dx) and then solve for the original integral algebraically when it reappears.
Sense of Study hint: Use LIATE to pick u, then set up a two-column table: u and dv on top, du and v on the bottom.
Worked Examples
Example 1
easySolution
- 1 LIATE: u = x, dv = e^x\,dx; then du = dx, v = e^x.
- 2 \int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C.
- 3 Factor: e^x(x-1) + C.
Answer
Example 2
hardExample 3
mediumPractice Problems
Try these problems on your own first, then open the solution to compare your method.
Example 1
easyExample 2
mediumRelated Concepts
Background Knowledge
These ideas may be useful before you work through the harder examples.