Question

Integration. 1/1+e 2x. 2x is the exponent properties

Answer 1

Well, my first thought is to do a little trick of adding and subtracting the same thing. Because if you add and subtract the same thing, you technically only add 0, which is legal to do. So I would do this:
\[\frac{ 1 }{ 1+e^{2x} }= \frac{ 1+e^{2x} -e^{2x} }{ 1+e^{2x} }\]Doing this allows me to make two fractions now like this:
\[\frac{ 1+e^{2x}-e^{2x} }{ 1+e^{2x} }= \frac{ 1+e^{2x} }{ 1+e^{2x} }- \frac{ e^{2x} }{ 1+e^{2x} }\]
Now from here that left fraction is just 1
\[\int\limits_{}^{}1dx -\int\limits_{}^{}\frac{ e^{2x} }{ 1+e^{2x} }dx\]We know the left integral. Now just do a u-sub in the second integral. Let u = 1+ e^(2x) and go from there.