$\int_0^{\infty} \frac{\ln x}{x^6 + 1}dx$ using contour integration I am looking for the solution to this improper integral. This was my approach:
Step I. (We use the 'pie slice' $\frac{\pi}{3}$). We define our log with $\frac{-\pi}{2} < \arg z < \frac{3\pi}{2}$.
\begin{equation}
\int_C \frac{\log(z)}{z^6 + 1}dz = \int_{L_1}\frac{\log(z)}{z^6 + 1}dz + \int_{L_2}\frac{\log(z)}{z^6 + 1}dz + \int_{\gamma_\epsilon}\frac{\log(z)}{z^6 + 1}dz + \int_{\Gamma_R}\frac{\log(z)}{z^6 + 1}dz
\end{equation}
Step II. Finding residues. 
The only singularity located within our contour is the singularity $e^\frac{\pi i}{6}$. 
\begin{equation}
2\pi i \operatorname{Res}(f; e^\frac{\pi i}{6}) = 2 \pi i\lim_{z\rightarrow z_0}\frac{(z-z_0)\log(z)}{z^6 - 1} = \frac{\pi^2}{18}e^\frac{\pi i}{6}
\end{equation}
Step III. Show that the integrals over $\Gamma_R$ and $\gamma_\epsilon$ $\rightarrow 0$ as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$. We omit this as it is not a necessary part for this question.
Step IV. Parametrise the integral over $L_2$.
\begin{equation}
\int_{L_2}\frac{\log z}{z^6 + 1}dz = -e^{\frac{\pi i}{3}}\int_{\epsilon}^{R} \frac{\ln(x) + i \frac{\pi}{3}}{x^6 + 1}dx
\end{equation} 
Step V. When putting all of this together, we have
\begin{align}
\frac{\pi^2}{18}e^\frac{\pi i}{6} = (1-e^{\frac{\pi i}{3}})\int_{0}^{\infty} \frac{\ln(x)}{x^6 + 1}dx - i 
 e^\frac{\pi i}{3}\int_{0}^{\infty}\frac{\frac{\pi}{3}}{x^6 + 1}dx
\end{align}
Which is unfortunate because the answer is not real and therefore certainly not correct. Can you see what I did wrong?

The formula you've obtained is correct. The formula contains imaginary units, but it's not yet the final answer. Except for assmuing that the formula is incorrect, you did nothing wrong. 
You have (after multiplying by $e^{-\frac{i \pi}{6}}$):
$$\frac{\pi^2}{18} = -i \int_{0}^\infty \frac{\ln x}{x^6+1}dx + \frac{\pi(1-i\sqrt{3})}{6} \int_0^\infty \frac{dx}{x^6+1} $$
Knowing that the integrals have real values, you can take the real and the imaginary part of this equation
$$\frac{\pi^2}{18} = \frac{\pi}{6} \int_0^\infty \frac{dx}{x^6+1} $$
$$0 = - \int_{0}^\infty \frac{\ln x}{x^6+1}dx - \frac{\pi\sqrt{3}}{6} \int_0^\infty \frac{dx}{x^6+1} $$
Solving this set of  equations gives you
$$ \int_0^\infty \frac{dx}{x^6+1} = \frac\pi 3$$
$$ \int_{0}^\infty \frac{\ln x}{x^6+1}dx = - \frac{\pi^2\sqrt{3}}{18}$$
If you want to obtain the values of these integrals without 'assuming' they are real, you can follow the method that Oscar Lanzi suggests in his answer.
