Integration by Parts Worksheet with Complete Solutions
Twenty problems covering polynomial products, logarithms, inverse trigonometric functions, definite integrals, and cyclic cases.
What is included
Practice choosing u and dv, repeated integration by parts, and cyclic integrals with a printable key.
Skills assessed
- LIATE judgment
- repeated integration by parts
- definite integral evaluation
Prerequisites
- antiderivatives
- product rule
Long description
Read the diagram from top to bottom. Each numbered box names one decision or mathematical operation. Arrows show the required order; the text labels remain the complete interpretation in print, dark mode, and nonvisual reading.
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Printable preview
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
- Evaluate .
Complete worked solutions
Every problem has a source-matched answer and independently reviewed derivation.
Problem 1: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 2: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 3: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 4: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 5: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 6: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 7: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 8: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 9: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 10: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 11: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 12: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 13: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification gives .
Problem 14: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Problem 15: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Problem 16: Evaluate .
Answer:
Why this method: Cyclic integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Problem 17: Evaluate .
Answer:
Why this method: Cyclic integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Problem 18: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Problem 19: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Problem 20: Evaluate .
Answer:
Why this method: Integration by parts matches the mathematical structure before any algebraic cleanup.
- Choose u as the factor that simplifies when differentiated and integrate dv.
- Apply ; repeat when the remaining integral still has a polynomial factor.
- Simplification and endpoint evaluation gives .
Common errors
- Choosing dv that is not readily integrable.
- Dropping the subtraction in the formula.
- Failing to solve for the original integral in a cyclic case.