How to evaluate the integral $\int_0^1\frac{\arctan x}{x}\frac{1-x^3}{1+x^3}dx$ I'm looking for the value of this integral:
$$\int_0^1\frac{\arctan x}{x}\frac{1-x^3}{1+x^3}dx$$
I try to integrate it:
\begin{align}
I&=\int_0^1\arctan x\left(\frac{1}{x}-\frac{2x^2}{1+x^3}\right)dx\\
&=\int_0^1\frac{\arctan x}{x}dx-2\int_0^1\frac{x^2\arctan x}{1+x^3}dx\\
&=G-\frac{2}{3}\left(\arctan x\ln (1+x^3)|_0^1-\int_0^1\frac{\ln (1+x^3)}{1+x^2}dx\right)\\
&=G-\frac{\pi}{6}\ln 2+\frac{2}{3}\int_0^1\frac{\ln (1+x)+\ln (x^2-x+1)}{1+x^2}dx\\
&=G-\frac{\pi}{6}\ln 2+\frac{2}{3}\frac{\pi}{8}\ln 2+\frac{2}{3}\int_0^1\frac{\ln (x^2-x+1)}{1+x^2}dx\\
&=G-\frac{\pi}{12}\ln 2+\frac{2}{3}\int_0^1\frac{\ln (x^2-x+1)}{1+x^2}dx\\
\end{align}
Where G is the Catalan's constant,but I don't know how to do it next .
 A: $$\color{blue}{I = \frac{2}{9}\pi \ln (2 + \sqrt 3 ) - \frac{{\pi \ln 2}}{{12}} - \frac{G}{9}}$$

Let $$I_1 = \int_0^1 \frac{\ln(1+x^3)}{1+x^2}dx \quad \quad I_2 = \int_1^\infty \frac{\ln(1+x^3)}{1+x^2}dx$$
Then subsituting $x\to 1/x$ in $I_1$ gives $$\tag{1} I_1 - I_2 = -3\int_1^\infty \frac{\ln x}{1+x^2} dx = -3G$$
Next we calculate $$I_1+I_2 = \int_0^\infty \frac{\ln(1+x^3)}{1+x^2} dx $$
Denote $$F(a) = \int_0^{ + \infty } {\frac{{\ln (a + {x^3})}}{{1 + {x^2}}}dx} $$
then, after some works, we obtain $$F'(a) = \int_0^{ + \infty } {\frac{1}{{(a + {x^3})(1 + {x^2})}}dx} = \frac{\pi }{2}\frac{a}{{1 + {a^2}}} + \frac{1}{6}\frac{{\ln a}}{{1 + {a^2}}} + \frac{{2\sqrt 3 \pi }}{9}\frac{{{a^{2/3}} - {a^{ - 2/3}}}}{{1 + {a^2}}}$$
Hence, after some calculations, $$\tag{2} I_1 + I_2 = F(0) + \int_0^1 {F'(a)da} = - \frac{G}{3} + \frac{{\pi \ln 2}}{4} + \frac{{2\pi }}{3}\ln (2 + \sqrt 3 ) $$
Combining (1) and (2) gives the value of $I_1$, and hence by what you have done
$$I = G-\frac{\pi}{6}\ln2 + \frac{2}{3} I_1 $$ we can find the value of $I$.
