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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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    there is no division in the problem statement...2011-03-15
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    I would expand it then integrate the usual way. Just a lot of distributive law. There could be a clever substitution, but I wouldn't bother.2011-03-15
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    Are you trying to integrate its reciprocal?2011-03-15
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    No, I was just trying to integrate as normal. My Calculus II class is currently at parts and trig substitutions so I was sure that one of the methods was applicable. I tend to use WolframAlpha to give me a starting point, however here it was really confusing when I saw "long division" in the step list.2011-03-15
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    the http://en.wikipedia.org/wiki/Binomial_theorem binomial theorem might help (so you dont multiply it all out the hard way!)2011-03-15
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    @chrisw: Ah, yes, Alpha does indeed say "by long division" when it expands that polynomial product. That's a bug in their "explanation" algorithm. You can safely ignore that.2011-03-15
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    @chrisw: I suspect the "intended answer" is to use trig substitution. See 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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    That would fit the stated context of the problem better!2011-03-15
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    @The Chaz: Precisely why I thought of trying it. It's so easy to raise $a+b$ to the fifth power, though, that on first brush I would be tempted to just expand and integrate. Now, if it was, say, the integral of $x^3(x^2+1)^{39}$ instead... (-:2011-03-15
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    This reminds me of the SAT mentality - make the exponent (or number of 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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    Thanks, this is what I ended up doing. I am still unsure why exactly WolframAlpha says to do long division?2011-03-15
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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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    Right around the time you posted this, I got an upvote *and a downvote* on my answer! Also, that's a pretty slick solution.2013-10-17
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    @TheChaz2.0 I guess this is the shortest one to do it.2013-10-17
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    +C, of course...2013-10-17