Solutions for an ODE I am looking for a solution of the ODE $x'(t)=x(t)+\frac{1}{1+e^{t}}$ which has finite limit when $t\rightarrow \infty$, I already find that the solutions are $e^t \ln(1+e^t)-te^t-1$  however these solutions do not converge as $t$ goes to infinity. Any idea? thanks.
 A: The limit of your solution 

$$e^t \ln(1+e^t)-te^t-1$$ 

as $t\to \infty$ exists and finite and equals to $0$.
A: Edit: Starting from you specific solution, you get that 
$$ e^t \ln(1+ e^t) - t e^t - 1 = e^t (\ln (1 + e^t) - \ln (e^t)) - 1 = \frac{\ln(e^{-t} + 1)}{e^{-t}} - 1 $$
which converges to $0$ as $t\to \infty$ through an application of L'Hopital. 

Original post follows:
First, I don't think your solution is quite correct 
Re-writing your equation as 
$$ x'(t) - x(t) = \frac{1}{1+\exp(t)} $$
we have that by integrating factors and a version of partial fractions we get
$$ e^t \frac{d}{dt}( e^{-t} x(t)) = \frac{1}{1+\exp(t)} \implies \frac{d}{dt} [\exp(-t)x(t)] = \frac{1}{\exp(t)} - \frac{1}{1 + \exp(t)} $$
The antiderivative on the right hand side can be explicitly evaluated
$$ \int e^{-t} = - e^{-t} $$
and
$$ -\int \frac{1}{1+ e^{t}} = \int \frac{- e^{-t}}{e^{-t} + 1} = \ln (1 + e^{-t}) $$
which gives us the solution, where $C$ is the constant of integration, 
$$ x(t) = C e^t - 1 + \frac{\ln (1 + e^{-t})}{e^{-t}} $$
As $t\to\infty$, the final term is indeterminate. We can apply L'Hopital's rule to it, which yields by
$$ \lim_{x\to 0^+} \frac{\ln(1+x)}{x} = \lim_{x\to 0^+} \frac{(1+x)^{-1}}{1} = 1 $$
that provided the constant of integration $C = 0$ (this, of course, depends on the initial condition), the function $x(t)$ satisfies $\lim_{t\to\infty} x(t) = 0$. 
