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.
 A: 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*}
A: Note
\begin{align}
&\int_0^\infty \frac{1-x(2-\sqrt x)}{1-x^3}dx=\int_0^\infty \frac{1+x^{\frac32}-2x}{1-x^3}dx\\= &\int_0^\infty \frac{1+x^{\frac32}}{1-x^3}dx
-\int_0^\infty \underset{x\to x^{\frac12} }{\frac{2x}{1-x^3}}dx= \int_0^\infty \frac{dx}{1-x^{\frac32}}
-\int_0^\infty \frac{dx}{1-x^{\frac32}}=0 
\end{align}
