definite integral $\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\frac{-1}{2}\ln 3$ 
Problem: Prove that $$\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\frac{-1}{2}\ln 3$$

My attempt:
I am able to prove : $$\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx=\int_1^\infty \frac {1-x}{(1+x^3)\ln x}dx.$$ I don't know if this relation is useful or not. By the way, it seems it's impossible to integrate the function directly. So, I am looking for an equation derived from integration by parts or other similar tricks but I can't go further.
Any help would be highly appreciated.
P.S: This problem was proposed by "Jalil Ghaderi" from Iran.

$$I=\int_0^1 \frac {1-x}{(1+x^3)\ln x}dx$$
Making change $x=e^{-t}$
$$I=-\int_0^\infty \frac {e^{-t}-e^{-2t}}{(1+e^{-3t})t}dt=-\int_0^\infty\frac{dt}{t}\big(e^{-t}-e^{-2t}\big)\big(1-e^{-3t}+e^{-6t}-+ ...\big)$$
$$=-\int_0^\infty\frac{dt}{t}\big(e^{-t}-e^{-2t}-e^{-4t}+e^{-5t}+e^{-7t}-e^{-8t}+-...\big)$$
$$=-\int_0^\infty\frac{dt}{t}\big(e^{-t}-e^{-2t}+e^{-3t}-e^{-4t}+- ...\big)+\int_0^\infty\frac{dt}{t}\big(e^{-3t}-e^{-6t}+e^{-9t}-+ ...\big)$$
$$=-\int_0^\infty\Big(\frac{e^{-t}}{1+e^{-t}}-\frac{e^{-3t}}{1+e^{-3t}}\Big)\frac{dt}{t}=-\int_0^\infty(f(t)-f(3t))\frac{dt}{t}$$
According to Frullani's theorem (https://en.wikipedia.org/wiki/Frullani_integral)
$$I=\big(f(0)-f(\infty)\big)\ln\frac{1}{3}=-\ln3\Big(\frac{1}{2}-0\Big)=-\frac{\ln3}{2}$$
