Note that 1/(2^x)=2^-x.

Also recall that d/dx(2^x)=log(2)*2^x

So our integral is -1/log(2) * 2^(-x)+C. We now this because when we take the derivative, we have a -log(2) to get rid of so that we will be left with 2^(-x) (evaluate between 0 and 1 of course).