Evaluating Integral $\int e^{x}(1-e^x)(1+e^x)^{10} dx$ I have this integral to evaluate: $$\int e^{x}(1-e^x)(1+e^x)^{10} dx$$
I figured to use u substitution for the part that is raised to the tenth power. After doing this the $e^x$ is canceled out.
I am not sure where to go from here however due to the $(1-e^x)$. 
Is it possible to move it to the outside like this and continue from here with evaluating the integral?
$$(1-e^x)\int u^{10} du$$
 A: If $u=1+e^x$, then $e^x=u-1$, so $1-e^x=1-(u-1)=2-u$. Now you can easily multiply out. Note that the ‘bad’ factor could be any simple polynomial in $e^x$, and the technique would still work. For instance, if the integrand had been $e^x(3e^{2x}-4e^x+5)(1+e^x)^{10}$, substituting $u=1+e^x$ would turn it into $(3u^2-2u+4)u^{10}$, since 
$$\begin{align*}
3e^{2x}-4e^x+5&=3(u-1)^2+4(u-1)+5\\
&=3u^2-2u+4\;.
\end{align*}$$
A: Let $u=e^x$, $du=e^xdx$, so your integral is
$$\int (1-u)(1+u)^{10}du.$$
Now there are many ways for you to proceed (the quickest probably being integration by parts).
Addendum: to prove my claim that the quickest way is by parts, here it is:
$$\int (1-u)(1+u)^{10}du = \int(1-u)d\left(\frac{(1+u)^{11}}{11}\right) =$$ 
$$(1-u)\left(\frac{(1+u)^{11}}{11}\right) + \int \frac{(1+u)^{11}}{11} du = \frac{(1-u)(1+u)^{11}}{11}+\frac{(1+u)^{12}}{132}+C.$$
I challenge anyone (cough! cough! :-)) to do it quicker by expanding $(1+u)^{10}$.
A: let $x=\ln(u)$
$dx=du/u$
$I=\int e^{x}(1-e^x)(1+e^x)^{10} dx$ = $\int ((u(1-u)(1+u)^{10})/u)du$=$\int (1-u)(1+u)^{10}du$ 
You may want to take it from here...
A: FWIFs, this also would be easily generalized via a recurrence relation (aside from the obvious substitution that confirms this] as the integral is of the form;
f ' (x) times g(x) (where f(x) is (1/11)(1+e^x)^11) and f(x), g(x) both behave good enough - i.e. we get back something close enough to our integral the define a recurrence.
