Can I simplify$\displaystyle \int \frac {e^t}{e^t+1}dt$ into $\displaystyle \int 1+e^t dt$
so that the antiderivative will be $\displaystyle x+e^t+C$
what are you doing?! are you reading the posts made here?
picking up where Mush left off: we have
$\displaystyle \int \frac 1u~du$
applying the rule i just gave you, we get
$\displaystyle \ln |u| + C$
now back-substitute for $\displaystyle u$, we get:
$\displaystyle \ln |e^t + 1| + C$
please re-read posts #3 and #5 carefully
Maybe some colour would help.
$\displaystyle {\color{red}u = e^t + 1} \ \Rightarrow \ {\color{blue}du = e^t dt}$
So your integral: $\displaystyle \int \frac{{\color{blue}e^t dt}}{{\color{red}e^t + 1}} = \int \frac{{\color{blue}du}}{{\color{red}u}}$
which is a standard one.