Solve this definite integral I have a problem with this definite integral. Please help me:
$$ f(x)= {\int }\frac{1}{1+e^{2x}} dx $$

HINT:
$$I=\int\frac1{1+e^{2x}}dx=\int\frac{e^{-x}dx}{e^{-x}+e^{x}}$$
Setting $\displaystyle e^{-x}=v,I=-\int\frac{vdv}{1+v^2}$
Set $1+v^2=u$
