Evaluating $\int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$ What methods would work best to find $\displaystyle \int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$
As usual with this kind of integral i tried to differentiate with the respect of a parameter
$$\int _0^1\frac{\ln \left(ax^3+1\right)}{x+1}\:dx$$
$$\int _0^1\frac{x^3}{\left(x+1\right)\left(ax^3+1\right)}\:dx=\frac{1}{a-1}\int _0^1\left(\frac{ax^2-ax+a}{ax^3+1}-\frac{1}{x+1}\right)\:dx$$
in the end $2$ of these integrals are nice but the other $2$ are not
$\displaystyle \int _0^1\frac{-ax}{ax^3+1}\:dx$ and $\displaystyle \int _0^1\frac{a}{ax^3+1}\:dx$
Is there a better approach to this?
 A: Instead of using Feynman's trick at once use the following substitution first
$$\underbrace{\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx}_{x=\frac{1-t}{1+t}}$$
$$=\ln \left(2\right)\underbrace{\int _0^1\frac{1}{1+x}\:dx}_{\ln \left(2\right)}+\int _0^1\frac{\ln \left(1+3x^2\right)}{1+x}\:dx-3\underbrace{\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx}_{\frac{1}{2}\ln ^2\left(2\right)}$$
Now make use of the general result proved here.
$$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x}\:dx$$
$$=-\frac{\ln ^2\left(b\right)}{4}-\frac{\text{Li}_2\left(-\frac{a}{b}\right)}{2}-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\ln \left(2\right)\ln \left(a+b\right)$$
So by setting $a=3$ and $b=1$ we have
$$-\frac{\text{Li}_2\left(-3\right)}{2}-\arctan ^2\left(\sqrt{3}\right)+\ln ^2\left(2\right)-\frac{1}{2}\ln ^2\left(2\right)$$
Thus,
$$\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}+\frac{1}{2}\ln ^2\left(2\right)$$
Where $\text{Li}_2\left(z\right)$ is the Dilogarithm function.
As suggested by others one could also turn the integral into
$$\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx=\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx+\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx$$
From the result above we obtain the value of that other integral which is
$$\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}$$
A: For once, Feynman's trick makes the problem more difficult.
Making the problem more general, I would write
$$I=\int _0^1\frac{\log \left(x^n+1\right)}{x+1}\,dx=\sum_{i=1}^n\int _0^1\frac{\log \left(x-r_i\right)}{x+1}\,dx$$ where the $r_i$ are the roots of unity. Now using
$$\int \frac{\log \left(x-r_i\right)}{x+1}\,dx=\text{Li}_2\left(\frac{r_i-x}{r_i+1}\right)+\log \left(\frac{x+1}{r_i+1}\right) \log   (x-r_i)$$
$$\int_0^1 \frac{\log \left(x-r_i\right)}{x+1}\,dx=\text{Li}_2\left(\frac{r-1}{r+1}\right)-\text{Li}_2\left(\frac{r}{r+1}\right)-$$ $$\log (-r) \log \left(\frac{1}{r+1}\right)+\log (1-r) \log
   \left(\frac{2}{r+1}\right)$$
