A surprising result

Since the degree of nominator of integrand $P(x) = 2x-3x^2-x^4$ is lower than the degree of denominator of integrand $Q(x) = x^6 + 2x^3+1 = (x^3+1)^2$ , we can apply Ostrogradsky's method as follows:

$\int \frac {P(x)}{Q(x)} dx = \frac {P_1(x)}{Q_1(x)} + \int \frac {P_2(x)}{Q_2(x)} dx$

where $Q_1 = \gcd (Q(x), Q'(x))$ and $Q_2 = Q(x)/Q_1(x)$ and $P_1(x)$ and $P_2(x)$ are polynomials one degree less than $Q_1(x)$ and $Q_2(x)$ respectively.

Since $Q'(x) = 6x^2(x^3+1)$ $\implies Q_1(x) = x^3 +1$ and $Q_2(x) = (x^3+1)^2 / (x^3+1) = x^3+1$ . We obtain $P_1(x)$ and $P_2(x)$ as follows.

$\begin{aligned} \int \frac {2x-3x^2-x^4}{x^6 + 2x^3+1} dx & = \frac {Ax^2+Bx+C}{x^3+1} + \int \frac {Dx^2+Ex+F}{x^3+1} dx \quad \quad \small \color{#3D99F6}{\text{Differentiate both sides}} \\ \frac {2x-3x^2-x^4}{x^6 + 2x^3+1} & = \frac {2Ax+B}{x^3+1} - \frac {3x^2(Ax^2+Bx+C)}{(x^3+1)^2} + \frac {Dx^2+Ex+F}{x^3+1} \\ 2x-3x^2-x^4 & = Dx^5 + (E-A)x^4 + (F-2B)x^3 + (D-3C)x^2 + (2A+E)x + B + F \end{aligned}$

Equating the coefficients, we get $A=C=1$ and $B=D=E=F=0$ . Therefore, we have:

$\begin{aligned} \int_0^1 \frac {2x-3x^2-x^4}{x^6 + 2x^3+1} dx & = \frac {x^2+1}{x^3+1} \bigg|_0^1 = \boxed{0} \end{aligned}$