Recipe

Integrate by parts with tabular integration

Use when: the same by-parts split has to be applied several times in a row, because one factor (usually a polynomial) needs repeated differentiation before it reaches $0$

Goal: the finished antiderivative in one pass, without hand-tracking u, v, du, dv through several separate applications

Taught in Repeated parts and tabular integration · extends Integrate by parts

Quick recipe

Build the D column
Build the I column
Multiply along the diagonals with alternating signs
Simplify and check

1
Put the polynomial factor (the one that would be $u$ in ordinary by-parts) at the top of a D column, then repeatedly differentiate it, one row per derivative, until it reaches $0$ .

Example

For $\int x^3e^{2x}\,dx$ : the D column is $x^3,\ 3x^2,\ 6x,\ 6,\ 0$ .

Why: this is the same repeated $u\to du\to d(du)\to\cdots$ chain that ordinary by-parts would produce one application at a time; the table just lays every round out at once.
2
Put the other factor at the top of an I column, then repeatedly antidifferentiate it the same number of times as the D column has rows.

Example

For the same integral, $dv=e^{2x}\,dx$ , so the I column is $e^{2x},\ \tfrac12e^{2x},\ \tfrac14e^{2x},\ \tfrac18e^{2x}$ .

Why: each I-column entry is exactly the $v$ that ordinary by-parts would compute at that round.
3
Starting at $+$ , multiply each D-row by the I-entry one row below it, alternating the sign every row, and stop once the D column hits $0$ .

Example

$x^3\left(\tfrac12e^{2x}\right)-3x^2\left(\tfrac14e^{2x}\right)+6x\left(\tfrac18e^{2x}\right)-6\left(\tfrac1{16}e^{2x}\right)+C$
- the D column reaches exactly $0$ after finitely many rows (a polynomial factor) — stop here; the table is complete.
- the D column never reaches $0$ (like $e^x$ paired with $\sin x$ ) — tabular integration in this simple form doesn't terminate; fall back to ordinary repeated by-parts and solve algebraically for the repeating integral instead.
Why: the alternating diagonal product is exactly the sum of boundary terms that repeated by-parts would have produced one round at a time, collected in a single pass.
4
Combine like terms, then differentiate the result to confirm it reproduces the original integrand.

Example

$\int x^3e^{2x}\,dx = \frac{x^3}{2}e^{2x}-\frac{3x^2}{4}e^{2x}+\frac{3x}{4}e^{2x}-\frac{3}{8}e^{2x}+C$
Differentiating regenerates $x^3e^{2x}$ after the other terms cancel in pairs.

Why: the same arithmetic-slip risk from ordinary by-parts is still present here, just spread across more rows.

Common mistakes

Forgetting to alternate the sign, or starting the alternation at $-$ instead of $+$ .
Stopping the table one row early, before the D column actually reaches $0$ .
Swapping which factor goes in the D column, so the "wrong" factor is the one being repeatedly differentiated.
Trying to force a table with no terminating row (like $e^x\sin x$ ) instead of recognizing that case needs the algebraic self-referential method instead.