By replacing $x$ with $\log t$ we have: $$\int \frac{dx}{e^x+1} = \int \frac{dt}{t(t+1)}=\log t-\log(t+1)\tag{1}$$ as well as: $$\int \frac{x\,dx}{e^x+1} = \int \left(\frac{1}{t}-\frac{1}{t+1}\right)\log t\,dt = \frac{1}{2}\log^2 t+\int\frac{\log t}{1+t}\,dt.\tag{2}$$ The dilogarithm makes his appearance now. We have: $$ -\log(1-z)=\sum_{n\geq 1}\frac{z^n}{n},\qquad \operatorname{Li}_2(z)=\sum_{n\geq 1}\frac{z^n}{n^2}=\int_{0}^{z}\frac{-\log(1-t)}{t}\,dt \tag{3}$$ that, together with integration by parts, gives: $$ \int\frac{\log t}{1+t}\,dt = \log t \log(1+t) + \operatorname{Li}_2(-t).\tag{3}$$ In a similar fashion, $$\operatorname{Li}_3(z)=\sum_{n\geq 1}\frac{z^n}{n^3}=\int_{0}^{z}\frac{\operatorname{Li}_2(t)}{t}\,dt \tag{4}$$ together with integration by parts leads to:
> $$\int\frac{x^2\,dx}{e^x+1}=\int\frac{\log^2 t}{t(t+1)}=\frac{\log^3 t}{3}-\log^2 t\log(t+1)-2\log t\operatorname{Li}_2(-t)+2\operatorname{Li}_3(-t).$$