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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?

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    You could try factorizing $x^3 + 1$. Although, I don't imagine all of the resulting integrals would be very "nice". – an4s Jul 26 '20 at 05:37

2 Answers2

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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 $$\int \frac{\log \left(x-r_i\right)}{x+1}\,dx=\text{Li}_2\left(\frac{r_i-x}{r_i+1}\right)+\log \left(\frac{x+1}{r_i+1}\right) \log (x-r_i)$$ $$\int_0^1 \frac{\log \left(x-r_i\right)}{x+1}\,dx=\text{Li}_2\left(\frac{r-1}{r+1}\right)-\text{Li}_2\left(\frac{r}{r+1}\right)-$$ $$\log (-r) \log \left(\frac{1}{r+1}\right)+\log (1-r) \log \left(\frac{2}{r+1}\right)$$

Claude Leibovici
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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 $$\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}$$

Dennis Orton
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