<s>[INST] 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.
 [/INST] 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}
</s>