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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?

$$(1-e^x)\int u^{10} du$$

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$1-e^x=2-(1+e^x)$ –  yoyo Dec 22 '11 at 0:41
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No, we cannot move it outside, it is not a constant. Let $u=1+e^x$. Then $du=e^x\,dx$ and since $e^x=u-1$, we find that $1-e^x=1-(u-1)=2-u$. –  André Nicolas Dec 22 '11 at 0:48
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4 Answers 4

up vote 3 down vote accepted

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

$$\begin{align*} 3e^{2x}-4e^x+5&=3(u-1)^2+4(u-1)+5\\ &=3u^2-2u+4\;. \end{align*}$$

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Thank you. You explained this pretty well. –  SineCosine Dec 22 '11 at 2:58
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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 $(1+u)^{10}$.

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The quickest way is just to do it. It's a polynomial after all. –  jspecter Dec 22 '11 at 1:01
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yeah. $ {}{}{} $ –  jspecter Dec 22 '11 at 1:13
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Pretty sure it's quicker by parts! I'll race you if you want. :) –  Bruno Joyal Dec 22 '11 at 1:25
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I think it will depend on whether or not you know the binomial formula. If you do, it's instantaneous. –  jspecter Dec 22 '11 at 1:28
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I do not think that word means what you think that it means. @jspecter –  The Chaz 2.0 Dec 22 '11 at 1:39
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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.

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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...

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