ok, integral of dx/(1 + 2e^3x)
let u = e^3x, so du = 3e^3x dx = 3udx
so dx = du/(3u)
substitute back in, you get the integral of du/(3u*(1 + 2u))
now there's an integral for this, integral of du/(u*(a + bu)) = (1/a)*ln(u/(a + bu)) + C
here, a = 1 and b = 2, and the whole thing is multiplied by 1/3
so you get (1/3)*ln(u/(1 + 2u)) + C
plug e^3x in for u, and the result is (1/3)*ln(e^3x/(1 + 2e^3x)) + C
just that easy!