Jonas Meyer has already pointed out a link where the properties of this integral are discussed. However, I would like to show you an alternative proof that this integral does not converge absolutely (I saw this in Analysis I/II by Zorich and it doesn't seem as well-known as it should be imho):
The idea is simple. We first see that the only issue with convergence is at $\infty$, since $\frac{\sin(x)}{x}$ is continuous at $0$.
Now $0\le |\sin(x)| \le 1$ implies
$$\frac{|\sin(x)|}{x} \ge \frac{\sin^2(x)}{x}$$
And we note that $\int_1^\infty \frac{\sin^2(x)}{x}dx$ should essentially have the same convergence-properties as $\int_1^\infty \frac{\cos^2(x)}{x}dx$ (with some hand-waving at this point). But if this is true, then $\int_0^\infty \frac{\sin^2(x)}{x}dx$ converges if and only if $$\int_1^\infty \left(\frac{\sin^2(x)}{x}+ \frac{\cos^2(x)}{x}\right) dx = \int_1^\infty \frac{dx}{x}$$
converges. The latter is clearly not true, so $\int_0^\infty \frac{\sin^2(x)}{x}dx$ doesn't converge, either.
More formally, we have
$$
\begin{align}
\int_0^\infty \frac{|\sin(x)|}{x} dx &\ge \int_{\pi/2}^\infty \frac{\sin^2(x)}{x} dx \\
&= \int_{0}^\infty\frac{\cos^2(x)}{x+\pi/2} dx \\
\end{align}
$$
Hence
$$
\begin{align}
\int_0^\infty \frac{|\sin(x)|}{x} dx &\ge \frac12 \int_0^\infty \left(\frac{\sin^2(x)}{x} + \frac{\cos^2(x)}{x+\pi/2} \right)dx \\
&= \frac12\int_{0}^\infty\left(\frac{1}{x+\pi/2} + \frac{\frac\pi2 \sin^2(x)}{x(x+\pi/2)} \right) dx \\
&\ge \frac12 \int_{0}^\infty\frac{1}{x+\pi/2} dx \\
&= \infty
\end{align}
$$
Best Answer
For conditional convergence note that
$$\int_1^c \frac{\sin x^{1/3}}{x- \ln x}dx = \int_1^{c^{1/3}} \frac{\sin u}{u^3- 3\ln u}3u^2 \, du = 3\int_1^{c^{1/3}} \frac{\sin u}{u- 3u^{-2}\ln u} \, du$$
The integral on the RHS is convergent by Dirichlet's test since $(u- 3u^{-2}\ln u)^{-1} \to 0$ monotonically as $u \to \infty$.
To prove that the integral is not absolutely convergent exploit the periodicity of $\sin$ by decomposing into a divergent sum of integrals over intervals of the form $[k\pi, (k+1)\pi]$. Proceed as is done for the well-known proof that $\int_1^\infty x^{-1} \sin x \, dx$ is not absolutely convergent.