Step 1: Substitution
Let $t = e^x$, so $dx = \frac{dt}{t}$. The limits go from $t = 1$ to $t = \infty$.
Step 2: Partial fractions
\[ \int_1^{\infty}\frac{dt}{t(1 + t)} = \int_1^{\infty}\left(\frac{1}{t} - \frac{1}{1 + t}\right)dt = \left[\ln\frac{t}{1+t}\right]_1^{\infty} \]
Step 3: Evaluate
At $\infty$, $\frac{t}{1+t} \to 1$ and $\ln 1 = 0$. At $1$, $\ln\frac{1}{2}$.
Step 4: Result
$0 - \ln\frac{1}{2} = \ln 2$. The other options contain $e$ or are negative, but the integrand is positive so the answer must be positive and equal to $\log 2$.
Final Answer:
The value is log 2. This is option (A).
\[ \boxed{\text{(A) }\log 2} \]