Solving $\int\frac{\ln(1+e^x)}{e^x} \space dx$ I'm trying to solve this integral.
$$\int\frac{\ln(1+e^x)}{e^x} \space dx$$
I try to solve it using partial integration twice, but then I get to this point (where $t = e^x$ and $dx = \frac{1}{t} dt$):
$$\int\frac{\ln(1+t)}{t^2} \space dt = \frac{1}{t} \cdot \ln(1+t) - \frac{1}{t} \cdot \ln(1+t) + \int\frac{\ln(1+t)}{t^2} \space dt$$
$$\cdots$$
$$0 = 0$$
What am I doing wrong?

Edit:
Your problem is having integrated by parts twice. Doing it one time looks like this:
$$\int \frac{\ln(1+t)}{t^2} \ \mathrm{d}t = -\frac1{t}\ln(1+t) - \int -\frac1{t}\cdot\frac1{1+t} \  \mathrm{d}t = -\frac1{t}\ln(1+t) + \int \frac1{t}\cdot\frac1{1+t} \  \mathrm{d}t$$
That new integral should be evaluated with partial fractions:
$$
\int \frac1{t(t+1)} \ \mathrm{d}t = \int \frac1{t}-\frac1{t+1} \ \mathrm{d}t = \ln t - \ln (t+1) + C
$$
If you want to check, the result is the following:
$$ \int \frac{\ln(1+e^x)}{e^x} \ \mathrm{d}x= x - \ln(1+e^x) - \frac{\ln(1+e^x)}{e^x} + C$$
