Trying to find the anti-derivative of $\tan x$ unsuccessfully by using Euler's formula

We know that
$$\tan t=\frac{e^{it}-e^{-it}}{i(e^{it}+e^{-it})}=\frac{e^{2i
    t}+1-2}{i(e^{2it}+1)}=-i(1-\frac{2}{e^{2it}+1}).$$
Now we try to find the anti-derivative of
$$
f(t)=\frac{1}{e^{2it}+1}.
$$
We observe that
$$
\frac{1}{e^{2it}+1}=\frac{1}{(e^{it}+i)(e^{it}-i)}=-2i(\frac{1}{e^{it}+i}-\frac{1}{e^{it}-i}).
$$
So we try to find the anti-derivative of
$$
\frac{1}{e^{it}+i},\frac{1}{e^{it}-i}.
$$
This is hard,so we give up this way.