How do I evaluate $\int \frac{\mathrm{d}x}{e^x + 1} $?

Setting $\displaystyle e^x=u,e^x\ dx=du\iff dx=\frac{du}{e^x}=\frac{du}u$$$\int\frac{dx}{e^x+1}=\int\frac{du}{u(u+1)}$$

Now $\displaystyle\frac1{u(u+1)}=\frac{u+1-u}{u(u+1)}=\frac1u-\frac1{u+1}$

HINT:

$$\frac1{e^x+1}=\frac{e^{-x}}{1+e^{-x}}$$ OR

$$\frac1{e^x+1}=1-\frac{e^x}{e^x+1} $$

It also equals $$1-\frac{e^x}{1+e^x}$$