What is the integral of $x(1-x)^8$? 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!
 A: *

*Method $1$: Use the binomial expansion of $(1-x)^{8}$ and then integrate term by term.

*Method $2$: Use integration by parts, by taking $u=x$ and $dv = (1-x)^{8} \ dx$. 
A: $$x(1-x)^8 = x-8 x^2+28 x^3-56 x^4+70 x^5-56 x^6+28 x^7-8 x^8+x^9$$
so
$$\begin{array}{cl}
&& \int x(1-x)^8 \,dx \\
&=& \int x-8 x^2+28 x^3-56 x^4+70 x^5-56 x^6+28 x^7-8 x^8+x^9 \,dx \\\\
&=& \int x \,dx - 8 \int x^2 \,dx +28 \int x^3 \,dx - 56 \int x^4 \,dx + 70 \int x^5 \,dx - 56 \int x^6 \,dx + 28 \int x^7 \,dx - 8 \int x^8 \,dx + \int x^9 \,dx \\
&=& \tfrac{1}{2} x^2 - \tfrac{8}{3} x^3 + \tfrac{28}{4} x^4 - \tfrac{56}{5}x^5  + \tfrac{70}{6} x^6 - \tfrac{56}{7} x^7 + \tfrac{28}{8} x^8 - 8 \tfrac{1}{9} x^9 + \tfrac{1}{10} x^{10} + C

\end{array}$$
A: $$\begin{align*}
\int x(1-x)^8\,dx &=\int x(x-1)^8\,dx\\
&=\int (x-1+1)(x-1)^8\,dx\\
&=\int ((x-1)^9+(x-1)^8)\,dx\\
&=\frac1{10}(x-1)^{10}+\frac19(x-1)^9+C\quad (C: \text{constant}).
\end{align*}$$
Note that in the last line I used
$$\int (x+a)^n\,dx=\frac1{n+1}(x+a)^{n+1}+C,\quad \text{when}\ n\ne -1.$$
A: Nobody seems to have mentioned what I'd have thought was an obvious point: It's simpler if the thing raised to the 8th power is just a variable rather than $1$ minus a variable, so let
$$
\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.)
