# Shortcut/trick for integrating a factored polynomial?

If I'm integrating a factored polynomial, say

$$\int{x(x+1)(x-2)(x+3)dx},$$

does some shortcut exist that keeps me from having to expand the polynomial? Currently, I'd just do all the multiplication which can get tedious with a higher order polynomial, and then integrate it once I had it in $ax^n$ form. I seem to remember learning some trick that helped here, and I saw this done very quickly in person a few weeks ago. I've looked on the web and text books but I haven't found anything.

Anyone know if such a thing exists?

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A $u$ sub is useless here as well; notice if we let $u=x+a$ for some $a$, it just makes the problem reappear. e.g. $u=x+1$ results in $\int x(x+1)(x-2)(x+3)dx=\int (u-1)u(u-3)(u+2)dx.$ –  000 Dec 5 '12 at 2:38
I think that integration by parts is the best way to go. However, there is a shortcut for differentiating factored polynomials. –  Akiva Weinberger Aug 24 '14 at 5:02

Note that the polynomial being integrated has no constant term. In particular, the lowest degree of any term is one: namely, $x(1)(-2)(3) = -6x$. Therefore, the antiderivative will have an $x^2$ term that can be pulled out.

Unfortunately, there is no reason to believe the rest of the antiderivative will be reducible. Indeed, in this case the antiderivative is:

$$\frac{1}{30}x^2 (6x^3 + 15x^2 - 50x - 90) + C$$

where the parenthetical cubic is irreducible over $\mathbb{Z}$. In other words, just because the original polynomial factored nicely, doesn't mean finding roots of its antiderivative will be easy.

You might now wonder, since we are integrating a fourth degree polynomial, if there is a way to know quickly what the coefficients will be of its fifth degree antiderivative. Note, however, that having the coefficients of this antiderivative is equivalent (in difficulty) to having the coefficients of the original polynomial. For example, the coefficient of $x^5$ in the antiderivative is $1/5$, so we conclude that the coefficient of $x^4$ in the polynomial pre-integration must have been $1$. In this way, knowing the expanded antiderivative would allow us to "recover" the coefficients of the factored polynomial we began with, and so having a "trick" to do the former would give us a "trick" to do the latter. Thus, the problem of integrating a factored polynomial by "trick" is basically equivalent in difficulty to expanding a polynomial by "trick."

Hopefully this gives you some intuition as to why there is no fast way to find said antiderivative.

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There are no cubic irreducible polynomials over $\mathbb R$: every cubic has a real root. –  lhf Dec 5 '12 at 4:09
Whoops, meant $\mathbb{Z}$ (not $\mathbb{R}$). Thanks. –  Benjamin Dickman Dec 5 '12 at 6:34
This makes perfect sense. I think my memory confused the fact that differentiating a factored polynomial is easy without doing the expansion with integration. When I saw someone zip through one it was part of a longer example, so they had most likely already computed the integral. Thanks for the clear explanation though! –  ccap Dec 5 '12 at 19:08

The only possible way that I have ever seen this being done is by recursively using integration by parts, very similar to the argument above.

I am not sure you want to go this way. If the roots of a polynomial are known, so are its coefficients (taking care of a multiplicative constant). They may be a bit hard to calculate by hand, but are an absolute piece of cake when coded. And integrating polynomials, given coefficients, is the easiest thing you can ask for.

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Okay, I just wrote out a really long and useless answer involving the extended product rule and an extension of integration by parts. Turns out, that's absolutely ugly and useless here. Let's use the traditional integration by parts.

Let $u'(x)=x(x+1)$ and $v(x)=(x-2)(x+3).$ Integrating $u'(x)$, we have $$\int x(x+1)dx=\int x^2+xdx=\frac{x^3}{3}+\frac{x^2}{2}.$$

Taking the derivative of $v(x)$ via the product rule, we have $$v'(x)=(x+3)+(x+2).$$

As a result of integration by parts,

$$\int u'vdx=uv-\int uv'dx=\left(\frac{x^3}{3}+\frac{x^2}{2}\right)(x-2)(x+3)-\int \left(\frac{x^3}{3}+\frac{x^2}{2}\right)\left((x+3)+(x+2) \right)dx$$

This is somewhat less intense and it is not as mistake prone. Do you feel this helps? I will try to think of a more clever idea.

## Observation of factorization ? ? ?

\begin{align} \int x(x+1)(x-2)(x+3)dx&=\int x^2(x-2)(x+3)dx+\int x(x-2)(x+3)dx\\ &=\int x^3(x+3)dx-2\int x^2(x+3)dx\\ &+\int x^2(x+3)dx-2\int x(x+3)dx\\ &=\int x^3(x+3)dx-\int x^2(x+3)dx-2\int x(x+3)dx . . . \end{align}

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