Your substitution is good:
$\displaystyle u=e^x+1\,\therefore\,du=e^x\,dx$ giving us:
$\displaystyle \int_2^{e+1} \frac{1}{u}\,du$
Can you proceed from here?
Incidentally, the given answer is not quite correct.
By the way, I would post integration problems in the calculus forum.
when a substitution is done, one may "reset" the limits of integration from x-values to u-values ...
$\displaystyle u = 1 + e^x$
for $\displaystyle x = 0$ , $\displaystyle u = 1 + e^0 = 2$
for $\displaystyle x = 1$ , $\displaystyle u = e^1 + 1 = e + 1$
therefore ...
$\displaystyle \int_0^1 \frac{e^x}{1+e^x} \, dx = \int_2^{e+1} \frac{1}{u} \, du = \left[\ln{u}\right]_2^{e+1} = \ln(e+1) - \ln(2) = \ln\left(\frac{e+1}{2}\right)$
I don't know what your book gave but you posted the answer as $\displaystyle ln\left(\frac{1+ e^x}{2}\right)$. Since this is a definite integral there should be no "x" in the answer. The correct answer is, as skeeter said, $\displaystyle ln\left(\frac{1+ e}{2}\right)$