Dogbone contour integral/branch cuts/residue at infinity

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\root{x - x^{2}} \over x + 2}\,\dd x:\ {\large ?}}$

${\large\tt\mbox{Following my own above comment:}}$

\begin{align} &\color{#c00000}{\int_{0}^{1}{\root{x - x^{2}} \over x + 2}\,\dd x} =\int_{\infty}^{1} {\root{1/x - 1/x^{2}} \over 1/x + 2}\,\pars{-\,{\dd x \over x^{2}}} =\int_{1}^{\infty} {\root{x - 1} \over 1 + 2x}\,{\dd x \over x^{2}} \\[3mm]&=\half\int_{0}^{\infty} {\root{x} \over \pars{x + 3/2}\pars{x + 1}^{2}}\,\dd x \\[3mm]&=\half\braces{2\pi\ic\bracks{ \overbrace{{\root{3/2}\expo{\ic\pi/2} \over \pars{-3/2 + 1}^{2}}} ^{\ds{=\ 2\root{6}\ic}}\ +\ \overbrace{\lim_{z \to \expo{\ic\pi}}\totald{}{z}\pars{\root{z} \over z + 3/2}} ^{\ds{=\ -5\ic}}}} -\half\int_{\infty}^{0} {\root{x}\expo{\ic\pi} \over \pars{x + 3/2}\pars{x + 1}^{2}}\,\dd x \end{align}