Young woman in a thoughtful pose standing in front of a blackboard that is covered in equations

Integration techniques: Integration by Parts

Integration is the reverse of differentiation, but is quite a lot more frustrating for most people. This is the first of a series of articles on integration techniques.

Integration by Parts is basically the chain rule for integration, used when you have a product of two functions inside of your integral, like ∫3x2sin(x) dx  or  ∫x·ex dx.

The formula for integration by parts is given by:

u dv = uv – ∫v du


  • u is a function that you choose
  • dv is the differential of another function that you choose

then you calculate:

  • du is the derivative of u
  • v is the integral of dv

Step 1: Choose u and dv

(This is the hard part!)

You want to choose a u that is easy to differentiate and a dv that is easy to integrate. There may be a bit of trial-and-error involved in this part, so be patient with yourself, and allow plenty of time to work of these sorts of problems.

Tip: A good way to choose u is to go down the list following functions arranged in order of priority:

  1. logarithmic
  2. inverse trigonometric
  3. algebraic (AKA a polynomial)
  4. trigonometric
  5. exponential

So, if you have both an algebraic and a trigonometric function, you would choose the algebraic function as your u. After you choose what u is, everything left is your dv.

(This handy list to simplify your choice doesn’t seem to be in any of the standard calculus textbooks I have on hand, but was included in Calculus for Dummies by Mark Ryan, page 266, ©2014, John Wiley & Sons, Inc., where it, in turn, cites an article by Herbert E. Kasube in the American Mathematical Monthly 90, 1983 issue, and suggests the acronym LIATE, or Let’s Integrate Another Tantalizing Example, to help remember the order.)

Step 2: Calculate du and v

Now that you have chosen u and dv, calculate the derivative du of u, and the integral v of dv.

\\ \: \\ \large \mathrm{{\color{orangered} du \color{black} = \frac{d}{dx} \color{red}u}} \\ \: \\ and \\ \: \\ \large \mathrm{\color{purple}v \color{black} = \int{ \color{blue}dv}} \\ \: \\

Step 3: Apply the Integration by Parts Formula

You now have the four pieces you needed to find in order to use the Integration by Parts formula. Plug u, du, v, and dv into the formula, and what you get should be much easier to solve than the original integral.

\\ \: \\ \large \mathrm{\int{\color{red} u \color{blue}dv} \color{black} = \color{red}u \color{purple}v \color{black} - \int{\color{purple}v \color{orangered}du}} \\ \: \\

Step 4: Solve the New Integral

You now have a new integral on the right side of the equation, which hopefully is simpler than the original one. Solve it if possible. (If you still have products in functions in the new integral, you might need to go through all these steps again for the new one, repeating until you reach one that you can solve.)

Step 5: Remember the Constant of Integration

After writing down your final answer, always remember to tack ” + C ” for the unknown constant to the end of your answer. It’s easy to be so relieved that you reached an answer that you forget this important last step.

And … you are finished!

Be sure to check out examples one and two next, as well as the pages on integration by substitution and integration by partial fractions! If you have questions about this or think of anything you would like to see covered, please leave a comment below! 🙂