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I am trying to integrate $x^3(x^2+1)^5$ and I believe the starting point is to utilize long division to break up the polynomial, however I am unsure how to divide this out. Could someone give me what exactly I'm suppose to be dividing?

Edit: Well the reason I ask is if you attempt to use wolframalpha.com, they state to use long division to simplify the problem and I don't see how that's possible. My first thought was it was a parts problem.

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    @chrisw: I suspect the "intended answer" is to use trig substitution. $S$ee my answer below.2011-03-15

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I would use integration by parts. Technically you need to figure out the factors of this polynomial to know what possible parts are, though you don't actually need polynomial division to do that.

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You can also use a trig substitution here. Let $x=\tan(\theta)$. Then $x^2 + 1 = \tan^2\theta + 1 = \sec^2\theta$, $x^3 = \tan^3\theta$, and $dx = \sec^2\theta\,d\theta$. You get \begin{align*} \int x^3(x^2+1)^5\,dx &= \int \tan^3\theta\sec^{12}\theta\,d\theta\\ &= \int \tan\theta(\sec^2\theta-1)\sec^{12}\theta\,d\theta\\ &= \int\tan\theta\sec\theta \sec^{13}\theta\,d\theta - \int\tan\theta\sec\theta\sec^{11}\theta\,d\theta\\ &= \frac{1}{14}\sec^{14}\theta - \frac{1}{12}\sec^{12}\theta + C\\ &= \frac{1}{14}(1+x^2)^7 - \frac{1}{12}(1+x^2)^6 + C. \end{align*}

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    This reminds me o$f$ the SAT mentality - make the exponent (or number o$f$ iterations) high enough that the student will waste a bunch of time if they do it the "hard" way, but low enough that they don't immediately start looking for an "easy" way!2011-03-15
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It's actually not that bad if we just expand the binomial and multiply through by $x^3$.

$\int {x^3(x^2+1)^5dx} = \int {x^3(x^{10} + 5x^8 + 10x^6 + 10x^4 + 5x^2 + 1)dx} =$

$ \int {(x^{13} + 5x^{11}+10x^9+10x^7+5x^5+x^3)dx}$

And now you can integrate term by term.

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    Yeah... maybe a link to a screenshot or a scan from the book would enlighten us?2011-03-15
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\begin{align} \int x^{3}\left(x^{2} + 1\right)^{5}\,{\rm d}x &= \int x\left[\left(x^{2} + 1\right)^{6} - \left(x^{2} + 1\right)^{5}\right] \,{\rm d}x = {\left(x^{2} + 1\right)^{7} \over 14} - {\left(x^{2} + 1\right)^{6} \over 12} \end{align}

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    +C, of course...2013-10-17