TITLE: What is the integral of $x(1-x)^8$?
QUESTION [7 upvotes]: I want to find the integral of $x (1-x)^8$. How do I go about this? For example, which rule do I use from http://integral-table.com ? Thanks!

REPLY [12 votes]: Nobody seems to have mentioned what I'd have thought was an obvious point: It's simpler if the thing raised to the 8th power is just a variable rather than $1$ minus a variable, so let
$$
\begin{align}
u & = 1-x, \\  \\
du & = - dx, \\  \\
x & = 1-u.
\end{align}
$$
Then
$$
\int x (1-x)^8\;dx = \int (1-u)u^8\;(-du) = \int u^9-u^8\;du.
$$
It's easy to antidifferentiate that, and then put $1-x$ wherever $u$ appears.
(After that, the 10th and 9th powers of $1-x$ can be expanded if desired.)