Math 129 Section 005H Lecture 3: Integration by parts (continued)¹¹ 1 This document is licensed under a Creative Commons Attribution 3.0 United States License Your turn

Tabular (optional)

(Friday, August 27, 2021
Revised August 29, 2021)

Tabular integration is useful for repeated integration by parts where $u(x)$ is a polynomial. For example:

\int x^{5}e^{2x}~{}dx

can be done by putting all the $u$ ’s in one column and all the $v^{\prime}$ s in another, like this:

	$u$	$v^{\prime}$
		$e^{2x}$
+	$x^{5}$	$\frac{1}{2}e^{2x}$
-	$5x^{4}$	$\frac{1}{4}e^{2x}$
+	$20x^{3}$	$\frac{1}{8}e^{2x}$
-	$60x^{2}$	$\frac{1}{16}e^{2x}$
+	$120x$	$\frac{1}{32}e^{2x}$
-	$120$	$\frac{1}{64}e^{2x}$

In the first column we put alternating $\pm$ signs. Notice how we line up the two columns. Then the antiderivative is

		$\displaystyle x^{5}\cdot\frac{1}{2}e^{2x}$
	$\displaystyle-$	$\displaystyle 5x^{4}\cdot\frac{1}{4}e^{2x}$
	$\displaystyle+$	$\displaystyle 20x^{3}\cdot\frac{1}{8}e^{2x}$
	$\displaystyle-$	$\displaystyle 60x^{2}\cdot\frac{1}{16}e^{2x}$
	$\displaystyle+$	$\displaystyle 120x\cdot\frac{1}{32}e^{2x}$
	$\displaystyle-$	$\displaystyle 120\cdot\frac{1}{64}e^{2x}+C$

=\Big{(}\frac{1}{2}x^{5}-\frac{5}{4}x^{4}+\frac{5}{2}x^{3}-\frac{15}{4}x^{2}+% \frac{15}{4}x-\frac{15}{8}\Big{)}e^{2x}+C.