How to solve $\int_{0}^{1}e^{x}\ln(1+e^{x})dx$? I am trying to solve this integral: 
$\int_{0}^{1}e^{x}\ln(1+e^{x})dx$
Here is my attempt:
$[e^{x}\ln(1+e^{x})]_{0}^{1} - \int e^{x} \frac{e^{x}}{1+e^{x}}dx$ 
$[e^{x}\ln(1+e^{x})]_{0}^{1} - \int e^{x} \frac{e^{x}}{1+e^{x}}dx$ 
$[e^{x}\ln(1+e^{x})]_{0}^{1} - ( \ln(1+e^{x})e^{x} - \int \ln(1+e^{x})e^{x}dx )$ 
Then I got the original integral back, my attempt was to then use this method:
$I = \int_{0}^{1}e^{x}\ln(1+e^{x})dx$
$I = [e^{x}\ln(1+e^{x})]_{0}^{1} -  \ln(1+e^{x})e^{x} + I$ 
$0 = [e^{x}\ln(1+e^{x})]_{0}^{1} -  \ln(1+e^{x})e^{x}$ 
And now I don't know what to do because the $I$ on the left and right cancel each other. Any help is greatly appreciated.

Consider the substitution
$$u= 1+e^x.$$
In this case, you have $dx = e^{-x} du$, $u=2$ as $x\to 0$, $u\to 1+e$ as $x\to 1$ and you integral turns into
$$ \int_2^{1+e} \log(u) du,$$
which can be easily solved using integration by parts.
To be more clear, integration by parts is:
$$\int_2^{1+e} f(u) g^{\prime}(u)\ du = f(u) g(u)\Big|_{2}^{1+e}-\int_2^{1+e} f^{\prime}(u) g(u) du.$$
Here you can take $f(u) = \log(u)$ and $g^{\prime}(u) =1.$
Then it follows that
$$ \int_2^{1+e} \log(u) du = u \log(u)\big|_2^{1+e} -\int_2^{1+e} 1 du =(u \log(u) - u)\Big|_2^{1+e}.$$
Going back with the substitution you have
$$\int_0^1 e^x \log(1+e^x) dx = \left[(1+e^x) \log(1+e^x) - 1-e^x)\right]\Big|_0^1 \\= (e+1)\log(e+1)-2\log(2) -e+1.$$
$\bf{EDIT:}$ I changed the extremals of the integral, but actually in the end I preferred to switch to the original integral to substitute the extremals.
