= Solution
\b[Poles and residues.] An isolated singularity $z_0$ is a <pole> of order $m$ if its <Laurent series> has finitely many negative powers, the lowest being a nonzero multiple of $(z-z_0)^{-m}$. The <residue> is the coefficient of $(z-z_0)^{-1}$.
If $h(z_0)=0$ and $h'(z_0)\ne0$, factor $h(z)=(z-z_0)H(z)$, where $H$ is <holomorphic> near $z_0$ and $H(z_0)=h'(z_0)$. For $g(z_0)\ne0$,
$$
\frac{g(z)}{h(z)}=\frac1{z-z_0}\frac{g(z)}{H(z)}.
$$
The last quotient is <holomorphic> and nonzero at $z_0$, proving a <simple pole> with
$$
\boxed{\operatorname{Res}_{z_0}\frac gh=\frac{g(z_0)}{h'(z_0)}.}
$$
\b[The given function.] For noninteger $a$, specify a <branch of a multivalued function> before using $z^a$. Choose $z^a=\exp(a\log z)$ with $-\pi/2<\arg z<3\pi/2$, a cut along the negative imaginary axis. This leaves all three roots of $1+z^3$ off the cut. They are $z_+=e^{i\pi/3}$, $z_- =e^{-i\pi/3}$ and $z_0=-1$, all simple. The numerator is nonzero at these points. Since the denominator has <derivative> $3z^2$, the <residues> are
$$
\boxed{\begin{array}{c|c}
\text{pole}&\text{residue}\\\hline
e^{i\pi/3}&\tfrac13e^{i\pi(a-2)/3}\\
e^{-i\pi/3}&\tfrac13e^{-i\pi(a-2)/3}\\
-1&\tfrac13e^{i\pi(a-2)}
\end{array}.}
$$
For noninteger $a$, zero is a <branch point>, rather than an isolated <pole>. This remains true when $-1<a<0$ and the numerator diverges there. For nonnegative integer $a$ it is a regular point. Other logarithm branches change the phases of the <residues> consistently.
\b[Integral and convergence.] Near zero the integrand behaves as $x^a$, requiring $a>-1$. At infinity it behaves as $x^{a-3}$, requiring $a<2$. Therefore the exact convergence range is
$$
\boxed{-1<a<2.}
$$
Use a counterclockwise sector contour with rays at angles $0$ and $2\pi/3$, inner radius $\varepsilon$ and outer radius $R$. The same <complex logarithm> branch is valid throughout this sector. Only $z_+=e^{i\pi/3}$ lies inside. The small circular arc contributes $O(\varepsilon^{a+1})$ and the large one contributes $O(R^{a-2})$, so both vanish precisely in the stated range.
The outward positive-real ray contributes $I$. On the returning ray set $z=xe^{2\pi i/3}$: the denominator is still $1+x^3$, while $z^a dz=e^{2\pi i(a+1)/3}x^a dx$, with reversed limits. The <residue theorem> gives
$$
(1-e^{2\pi i(a+1)/3})I=\frac{2\pi i}{3}e^{i\pi(a-2)/3}.
$$
Put $\alpha=\pi(a+1)/3$. The right-hand phase is $-e^{i\alpha}$ and $1-e^{2i\alpha}=-2ie^{i\alpha}\sin\alpha$. Hence
$$
\boxed{\int_0^\infty\frac{x^a}{1+x^3}\,dx=\frac{\pi}{3\sin(\pi(a+1)/3)},\qquad-1<a<2.}
$$
The sine is positive in this range, consistent with a positive real <integral>. This is a special case of the <Mellin integral of a reciprocal binomial>.
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