Integral $\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$ vanishes Come across
$$I=\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx$$
and break the integrand as
$$\frac{1-x(2-\sqrt x)}{1-x^3}=\frac{1-x}{1-x^3}- \frac{x(1-\sqrt x)}{1-x^3}
$$
The first term simplifies and the second term transforms with $\sqrt x\to x$. Then, the integral becomes
$$I=\int_0^\infty \frac{1}{1+x+x^2}dx
- 2\int_0^\infty \frac{x^3}{1+x+x^2+x^3+x^4+x^5}dx
$$
The first integration is straightforward $\frac{2\pi}{3\sqrt3}$ and the second is carried out via partial fractional decomposition. After some tedious and lengthy evaluation, it surprisingly yields the same value $\frac{2\pi}{3\sqrt3}$.
Given its vanishing value, there may exist a shorter and more direct derivation, without evaluation of any explicit intermediate values.

I'm not sure if this is something that is of value to you since it uses your approach, but once you transform $\sqrt{x} \to x$ for the second integral you can use PFD to cancel out the first integral and be left with two simple integrals, instead of expressing it as you did:
$$-\int_0^{\infty} \frac{2x^3(1-x)}{1-x^6} \; \mathrm{d}x=-\int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\frac{1}{3}\int_0^{\infty} \frac{2x-1}{x^2-x+1} \; \mathrm{d}x+\frac{2}{3} \int_0^{\infty} \frac{\mathrm{d}x}{x+1} $$
And so you have,
\begin{align*}
I&= \int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\int_0^{\infty} \frac{1}{x^2+x+1} \; \mathrm{d}x-\frac{1}{3}\int_0^{\infty} \frac{2x-1}{x^2-x+1} \; \mathrm{d}x+\frac{2}{3} \int_0^{\infty} \frac{\mathrm{d}x}{x+1} \\
&= \bigg [-\frac{1}{3} \ln(x^2-x+1)+\frac{2}{3}\ln(x+1) \bigg]\bigg \rvert_0^{\infty} \\
&=0
\end{align*}
