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6.11 Integrating Using Integration by Parts

3 min readβ€’june 18, 2024


AP Calculus AB/BC ♾️

279Β resources
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AP Description & Expectations 🎯

  • Enduring Understanding FUN-6: Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.Β 
  • Essential Knowledge FUN-6.E.1: Integration by parts is a technique for finding antiderivatives.


What is Integration by Parts?

Integration by parts is a method of integration that transforms an integral of a product of functions into an integral of the product of one function’s derivative and the others antiderivative. Although this sounds confusing at first, we can build this concept from the product rule:Β πŸ₯‡
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Whenever there’s an integral where the integrand (the thing being integrated) is a product of two functions, one of the functions will be f(x) and the other will be g’(x). If other methods like u-substitution don’t work, try integration by parts. πŸ’―

Another Form of the Equation ❓

Sometimes, you might see the equation written like this:
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This is the same concept in a different form.
One of the functions will be U and the other will be V. However, we still need to know what dV and dU actually mean. If U is a function of x, then U = f(x). Taking the derivative of both sides, we find that dU = f’(x)dx. Same thing for V: If V = g(x),Β  then dV = g’(x)dx. This new way of writing the integration by parts formula is just a sneakier version of the original formula. 🀭

How and Why Would I Use This Formula? πŸ€”

The best way to explain this formula is through an example. Take the following integral:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(572).png?alt=media&token=4f6238b4-8451-4825-b925-ebbd874acd54
At first glance, you may try u-substitution, but this method will take you nowhere. Instead, notice that the integrand is a product of two functions. This is an opportunity to use integration by parts. Let’s recall the formula:
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The first step should be to figure out what f(x) and g’(x) should be. This step is probably the most challenging of this method, because you have to choose which is which in order to make the problem work.Β 
This is where a handy acronym comes into play: L.I.A.T.E. It stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential.
The function closer to the β€œE” side should be g’(x) and the function closer to the β€œL” side should be f(x).Β 
For our example problem, that means:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(576).png?alt=media&token=40b70a38-9e35-421d-b317-64cbe13e0913

We know we need g(x) and f’(x) to use the integration by parts formula, so let’s find those now:
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Steps to Take for Integration by Parts

  1. Check to make sure that u-substitution and other methods don’t work.
  2. Check to make sure that the integrand is a product of two functions.
  3. Use L.I.A.T.E. to pick which function should be f(x) and which should be g’(x).
  4. Plug into the formula and integrate.

Note: You may have to use integration by parts more than once if your resulting integral is also a product of two functions that can't be solved using u-substitution.Β 

Worked Examples πŸ“”


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Browse Study Guides By Unit
πŸ‘‘Unit 1 – Limits & Continuity
πŸ€“Unit 2 – Fundamentals of Differentiation
πŸ€™πŸ½Unit 3 – Composite, Implicit, & Inverse Functions
πŸ‘€Unit 4 – Contextual Applications of Differentiation
✨Unit 5 – Analytical Applications of Differentiation
πŸ”₯Unit 6 – Integration & Accumulation of Change
πŸ’ŽUnit 7 – Differential Equations
🐢Unit 8 – Applications of Integration
πŸ¦–Unit 9 – Parametric Equations, Polar Coordinates, & Vector-Valued Functions (BC Only)
β™ΎUnit 10 – Infinite Sequences & Series (BC Only)
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