We study the integration technique of integration by parts.

### (Video) Calculus: Single Variable

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### Online Texts

- OpenStax II 3.1: Integration by Parts
- Ximera OSU: Integration by Parts
- Community Calculus 8.4: Integration by Parts

### Examples

Compute the indefinite integral

- Because integrating and differentiating are at a similar level of difficulty, we opt to differentiate so that its degree as a polynomial will be decreasing.
- This gives

Compute the indefinite integral

- Because integrating is much harder than differentiating it, we choose to differentiate and integrate .
- This gives

Compute the indefinite integral

- We’ll do this using an organizational technique called “tabular integration” that many people find helpful when doing repeated integrations by parts.
- Make a table: one column for and another for . In the first row, rewrite the
functions that you will use for and :
- Add new rows: differentiate items in the column and integrate items in the
column to determine what items go in the next row.
*one row down*. For the last item in the column, match it with the last item in the column (which you’ll end up using twice): - To finish, you alternate addition and subtraction. Give a to the first term, a to the second, and so on. Last but not least, put an integral on the last term as well: In this way, we arrive at the formula
- Sometimes the integrand of the last term is zero. This would happen when is a
polynomial if you have enough lines in your table. In that case there would be
nothing else to compute. (We stopped one line short of that point on purpose in
this example to show how you would handle a case where there the last term
*isn’t*zero.) - In this case, the final answer is

Compute the indefinite integral below using tabular integration:

- Let us take and (it would have been fine to choose them the other way around
as well). The table is then
Which gives

- This example has a twist: the integral on the right-hand side is just a constant
times the integral on the left-hand side. This indicates that further integrations
by parts would be unfruitful because you’d end up in something like a cycle.
What you
*can*do is*solve*for the answer by moving both integrals to the left side of the equation and then combining like terms: and therefore