$ I=\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}. $ The function on hand is \begin{gather*} f(z)=\frac{1}{z^{2/3}(1-z)^{1/3}} =\frac{e^{-\frac{2}{3}(\log |z|+ i (\arg z+2\pi k))}}{(1-z)^{1/3}}\\ =\frac{1}{z}\sqrt[\displaystyle3]{\frac{z}{1-z}} =\frac{e^{(\log|z|+ i (\arg z+2\pi k))/3}}{z(1-z)^{1/3}};\\ 0\le\arg z\le2\pi \; \; (\!\Rightarrow-1=e^{ i \pi}), \quad k=-1,0,1. \end{gather*}
Off the segment $\,[0,1]\,$ (or off any continuum containing the points $\,0,1$) this formula defines three different functions $\,f_{-1},f_0, f_1,\,$ - the branches of $\,f$. The points $\,0,1\,$ are third order branch points of $\,f.\,$ Circumventing each of these points changes the branch, returning to the initial one after the third rotation. Circumventing simultaneously the pair of the branch points keeps the branch unchanged, because the summary increase of the argument will be $\,\pm6\pi,\,$ which divided by $\,3\,$ amounts to $\,2\pi.\, $ Since passing from one branch to the next one occurs when only one of the branch points is circumvented, we only consider windings about $\,0,\,$ hence only $\,\arg z\,$ figures in the above expansions, and for which reason we're not interested in $\,\arg(1-z)\,$ (which value also doesn't change as a result of a full revolution around only $\,0$).
Hence the question of finding res$[f,\infty]\,$ has no meaning. What you can ask is finding the residue of the given fixed branch of the function.
The branches can be identified by their values at an arbitrary fixed point off the segment $\,[0,1],\,$ say, at $\,-1.\,$ We have, $\,f_k(-1)=2^{-1/3}e^{-2 i \pi(1+2k)/3}, k=-1,0,1.\,$ It's easy to see that $\,(C_r=\{|z|=r\})$, $ \int_{C_r}f_k(z)dz=\int_0^{2\pi}\sqrt[\displaystyle3] {\frac{re^{i\varphi}} {1-re^{i\varphi}}}\frac{i re^{i\varphi}\,d\varphi} {re^{i\varphi}} \to 2\pi i \{(-1)^{1/3}\}_k, \quad r \to \infty, $ and since the functions are holomorphic, taking the limit is the same as putting equality sign. Hence res$[f_k,\infty]=-e^{i\pi(1+2k)/3}, \; k=-1,0,1$.
Denoting by $\,I_k^-, I_k^+\,$ the integral over the segment $\,[0,1]\,$ for the values of $\,f_k\,$ accordingly on the lower and upper edges of the branch cut over $\,[0,1],\,$ we see that the value of the integral is the same for all three branches: \begin{gather*} 2\pi i e^{ i \pi(1+2k)/3}=\int_{C_r}f_k(z)\,dz=I_k^--I_k^+\\[-1.5ex] =\int_0^1\frac{dx}{x^{2/3}e^{4(1+k)\pi i /3}(1-x)^{1/3}} -\int_0^1\frac{dx}{x^{2/3}e^{4k\pi i /3}(1-x)^{1/3}}\\ =(e^{-4(1+k)\pi i /3}-e^{-4k\pi i /3})I =-e^{-4k\pi i /3}(1-e^{2 i \pi/3})I.\\ \Rightarrow I=-\frac{2\pi i e^{ i \pi(6k+1)/3}}{1-e^{2\pi i /3}} =-\frac{2\pi i }{e^{-\pi i /3}-e^{\pi i /3}} =\frac{\pi}{\sin\frac{\pi}{3}}; \\ \Rightarrow \boxed{ I = \frac{2\pi }{\sqrt{3}} \quad (k=-1,0,1)}. \end{gather*}
We may also use the following formula. Assume that $\,p\,$ is an integer and the rational function $\,Q\,$ has the finite poles $\,a_1,a_2,\ldots,a_n,\,$ none of which is zero or a positive number. Then \begin{equation} \int_0^\infty x^pQ(x)\,dx=-\sum_{k=1}^n{\rm res}[z^p{\rm Log} z\cdot Q(z); a_k], \tag{1} \end{equation} where $\,{\rm Log}\,$ is the branch of the logarithm, corresponding to $\,0\le\arg z\le2\pi\,$ (see multiple text-or-problem-books on Complex Analysis, say, the Russian edition (after second, 1972(?)) of the Сборник задач по ТФКП by L.I. Volkovyskii, G.L. Lunts, I.G. Aramanovitch (problems 4.169, 4.175), or the Spanish translation published in Moscow in 1977 (problems 711, 717). The Dover Publications edition, NY, 1965, ''A Collection of Problems on Complex Analysis'' is a translation from the first Russian edition and contains the incorrect versions of statements (problems 877, 883)). \begin{gather*} I=\int_0^1\sqrt[\displaystyle3]{\frac{x}{1-x}}\frac{dx}{x}. \\ {\rm Denote } \;\; t=\sqrt[\displaystyle3]{\frac{x}{1-x}}, \Rightarrow x=\frac{t^3}{1+t^3};\quad dx=\frac{3t^2}{1+t^3}; \; \Rightarrow I=3\int_0^{\infty}\frac{dt}{1+t^3} \end{gather*} We have $\,p=0\,$ in the situation of (1). The function $\,Q(z)=\frac{1}{1+z^3}\,$ has (simple) poles at $\,-1, e^{\pm i \pi/3}\,$ with residues $\,{\rm res}[{\rm Log}\cdot Q; -1]=\frac{ i \pi}{3},$ $\;{\rm res}[{\rm Log}\cdot Q; e^{ i \pi/3}]=\frac{ i \pi}{9}e^{-2 i \pi/3},$ $\; {\rm res}[{\rm Log}\cdot Q; e^{- i \pi/3}]=\frac{5 i \pi}{9}e^{2 i \pi/3}.\,$ According to (1): \begin{gather*} I=-3\{\frac{ i \pi}{3} +\frac{2 i \pi}{9}\frac{e^{-2 i \pi/3}+e^{2 i \pi/3}}{2} +\frac{4 i \pi}{9}e^{2 i \pi/3}\}\\ =-3\{\frac{ i \pi}{3}-\frac{2 i \pi}{9}\sin\frac{\pi}{6} +\frac{4 i \pi}{9}[-\sin\frac{\pi}{6}+ i \cos\frac{\pi}{6}]\} \Rightarrow \boxed{ I=\frac{2\pi}{\sqrt{3}}={\rm B}\left(\frac{1}{3},\frac{2}{3}\right)}. \end{gather*}