= Solution
Choose $\log u$ with $-\pi<\arg u<\pi$. The cut's upper and lower banks have $u=-x$, $x\in(0,1)$, and give
$$
i(-e^{i\pi z}+e^{-i\pi z})\int_0^1\frac{x^{z-1}}{x^2+4x+1}\,dx
=2\sin(\pi z)\int_0^1\frac{x^{z-1}}{x^2+4x+1}\,dx.
$$
The small circle about zero vanishes for $\operatorname{Re}z>0$. On the unit circle put $u=e^{i\theta}$. Since $u^2-4u+1=-2u(2-\cos\theta)$, its contribution is
$$
G(z)=\frac12\int_{-\pi}^{\pi}\frac{e^{i(z-1)\theta}}{2-\cos\theta}\,d\theta.
$$
Using $2-\cos\theta=1+2\sin^2(\theta/2)$ gives the required decomposition.
The only enclosed <pole> is $a=2-\sqrt3$; the other root $2+\sqrt3$ is outside. The <residue theorem>, including the factor $i$ in front of the contour integral, gives
$$
\boxed{I(z)=i(2\pi i)\frac{a^{z-1}}{2a-4}=\frac{\pi}{\sqrt3}(2-\sqrt3)^{z-1}}.
$$
Here the positive real logarithm defines the power, so this expression is an <entire function> of $z$. The finite-interval integral $G$ is also entire, by differentiation under the integral on compact sets. Therefore \b[$F(z)=\frac{\pi}{\sqrt3}(2-\sqrt3)^{z-1}-G(z)$] is an <analytic continuation> valid on the whole complex plane. Outside $\operatorname{Re}z>0$ it is this continuation, rather than the unregularized cut integral, that defines $F$.
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