58. Compute the integral

$$
\int_{-1}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}
$$
86.58. The main idea of the solution is to use the "non-evenness" of the function \( 1 /\left(1+x^{3}+\sqrt{1+x^{6}}\right) \), specifically:

\[
\begin{aligned}
& \text { of the function } 1 /\left(1+x^{3}+\sqrt{1+x^{6}}\right) \text {, specifically: } \\
& \int_{-1}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}=\int_{-1}^{0} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}+\int_{0}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}= \\
& =\int_{0}^{1} \frac{d x}{1-x^{3}+\sqrt{1+x^{6}}}+\int_{0}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}= \\
& =\int_{0}^{1}\left[\frac{1}{1-x^{3}+\sqrt{1+x^{6}}}+\frac{1}{1+x^{3}+\sqrt{1+x^{6}}}\right] d x= \\
& =\int_{0}^{1} \frac{\left(2+2 \sqrt{1+x^{6}}\right) d x}{\left(1+\sqrt{1+x^{6}}\right)^{2}-x^{6}}=\int_{0}^{1} 1 d x=1
\end{aligned}
\]