Shortcut/trick for integrating a factored polynomial?

Solution 1:

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.

Solution 2:

The trick is to recognize that the 4 roots of the polynomial, -3, -1, 0 and 2, are symmetric around $x=-\frac12$. Thus, with $x=t-\frac12$, the integral reduces to

$$\int{x(x+1)(x-2)(x+3)dx}=\int (t^2-\frac14)(t^2-\frac{25}{4})dt = \frac15t^5-\frac{13}6t^3 + \frac{25}{16}t $$

Solution 3:

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 ? ? ?

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