Solve this integral.. $$\int_{}^{}{\frac{x^2}{e^x+1}}dx$$
I have tried this using by-parts method but I could not do the integration of the second function.
 A: Set $e^x+1 \mapsto u$ then:
$$\int_{ }^{ }\frac{x^{2}}{e^{x}+1}dx=\int_{ }^{ }\frac{\left(\ln\left(u-1\right)\right)^{2}}{u\left(u-1\right)}du=\color{blue}{-\int_{ }^{ }\frac{\left(\ln\left(u-1\right)\right)^{2}}{u}du}+\color{red}{\int_{ }^{ }\frac{\left(\ln\left(u-1\right)\right)^{2}}{u-1}du}$$


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*For the red part setting $\ln\left(u-1\right) \mapsto t$ yields:
$$\int_{ }^{ }\frac{\left(\ln\left(u-1\right)\right)^{2}}{u-1}du=\int_{ }^{ }t^{2}dt=\frac{t^{3}}{3}+C=\color{red}{\frac{\left(\ln\left(u-1\right)\right)^{3}}{3}+C}$$



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*For the blue part use dilogarithm:


$$-\int_{ }^{ }\frac{\left(\ln\left(u-1\right)\right)^{2}}{u}du$$$$=-\left(\left(\ln\left(u-1\right)\right)^{2}\ln\left(u\right)-2\int_{ }^{ }\frac{\ln\left(u\right)\ln\left(u-1\right)}{u-1}du\right)$$$$=-\left(\left(\ln\left(u-1\right)\right)^{2}\ln\left(u\right)-2\left(-\text{Li}_{2}\left(1-u\right)\ln\left(u-1\right)+\int_{ }^{ }\frac{\text{Li}_{2}\left(1-u\right)}{u-1}du\right)\right)$$
Now calculate $\int_{ }^{ }\frac{\text{Li}_{2}\left(1-u\right)}{u-1}du$:
Set $1-u \mapsto v$:
$$\int_{ }^{ }\frac{\text{Li}_{2}\left(1-u\right)}{u-1}du=\int_{ }^{ }\frac{\text{Li}_{2}\left(v\right)}{v}du=\text{Li}_{3}\left(v\right)=\text{Li}_{3}\left(1-u\right)+C$$
$$\therefore \;\;\;-\int_{ }^{ }\frac{\left(\ln\left(1-u\right)\right)^{2}}{u}du=\color{blue}{-\left(\left(\ln\left(u-1\right)\right)^{2}\ln\left(u\right)-2\left(-\text{Li}_{2}\left(1-u\right)\ln\left(u-1\right)+\text{Li}_{3}\left(1-u\right)\right)\right)+C}$$

The final answer would be :
$$\int_{ }^{ }\frac{x^{2}}{e^{x}+1}dx$$

$$=\frac{\left(\ln\left(u-1\right)\right)^{3}}{3}-\left(\left(\ln\left(u-1\right)\right)^{2}\ln\left(u\right)-2\left(-\text{Li}_{2}\left(1-u\right)\ln\left(u-1\right)+\text{Li}_{3}\left(1-u\right)\right)\right)+C$$

