Find $A =\int \frac{x \, dx}{e^x +1 }$

Note that $$\begin{align} \int\,\frac{x\,\exp(x)}{\exp(x)+1}\,\text{d}x&=\int\,x\,\text{d}\ln\big(\exp(x)+1\big)\\&=x\,\ln\big(\exp(x)+1\big)-\int\,\ln\big(\exp(x)+1\big)\,\text{d}x\,. \end{align}$$ Then, we have that $$\int\,\ln\big(\exp(x)+1\big)\,\text{d}x=\int\,\frac{\ln\Big(1-\big(-\exp(x)\big)\Big)}{\big(-\exp(x)\big)}\,\text{d}\big(-\exp(x)\big)=-\text{Li}_2\big(-\exp(x)\big)+C\,,$$ where $C$ is a constant (as $\text{Li}_2(z)=-\int_0^z\,\frac{\ln(1-t)}{t}\,\text{d}t$, according to this). Thus, $$\begin{align}\int\,\frac{x}{\exp(x)+1}\,\text{d}x &=\int\,\left(x-\frac{x\,\exp(x)}{\exp(x)+1}\right)\,\text{d}x \\&=\frac{1}{2}x^2-x\,\ln\big(\exp(x)+1\big)-\text{Li}_{2}\big(-\exp(x)\big)+C\,.\end{align}$$

You can prove $\text{Li}_2(z)=-\int_0^z\,\frac{\ln(1-t)}{t}\,\text{d}t$ directly via $$\begin{align}\text{Li}_2(z)&=\sum_{k=1}^\infty\,\frac{z^k}{k^2}=\sum_{k=1}^\infty\,\int_0^z\frac{t^{k-1}}{k}\,\text{d}t=\int_0^z\,\frac{1}{t}\,\sum_{k=1}^\infty\,\frac{t^k}{k}\,\text{d}t \\ &=\int_0^z\,\frac{-\ln(1-t)}{t}\,\text{d}t\,, \end{align}$$ for all $z\in\mathbb{C}$ with $|z|<1$. Then, analytic continuation handles the rest.