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[Timothycok]

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Integral $\int_0^1\frac{\ln(1+x)}{x^2+x+1}\,dx$

$$I=\int_0^1\frac{\ln(1+x)}{x^2+x+1}\,dx$$
I tried $u$-substitution and $u=x+1$ but I didn't get any results. And I could not recognize this type of integration.

A:

$$\begin{align}I &= \int_{0}^{1}\frac{\ln(1+x)}{x^2+x+1}\,dx \\ &= \int_{0}^{1}\frac{\ln(1+x)}{x^2+x+1}\,dx - \int_{0}^{1}\frac{\ln(x)}{x^2+x+1}\,dx\\ &= \int_{0}^{1}\frac{\ln(1+x)}{x^2+x+1}\,dx - \int_{0}^{1}\frac{\ln(x)}{x^2+x+1}\,dx+\int_{0}^{1}\frac{\ln(x)}{x^2+x+1}\,dx\\ &= \int_{0}^{1}\frac{\ln(1+x)}{x^2+x+1}\,dx+\int_{0}^{1}\frac{\ln(x)}{x^2+x+1}\,dx-\int_{0}^{1}\frac{\ln(x)}{x^2+x+1}\,dx\\ &= \int_{0}^{1}\frac{\ln(1+x)}{x^2+x+1}\,dx+2\int_{0}^{1}\frac a77f14ba26 nekmai
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