Let f(x)=1/(a+be^(x)) and note that we can multiply the numerator (top) and denominator (bottom) of this fraction by e^(-x) to write f(x)=(e^{-x})/(ae^{-x}+b).
For the integral int (e^{-x})/(ae^{-x}+b)dx, use the substitution u=ae^{-x}+b, du=-ae^{-x}dx, and e^{-x}dx=-1/a du to obtain
int (e^{-x})/(ae^{-x}+b)dx=-1/a int\ 1/u du
=-1/a ln|u|+C=-(ln|ae^{-x}+b|)/a+C.
To see that this is the same as the second answer, note that
x=ln(e^{x})=ln|e^{x}| so that, by properties of logarithms,
x-ln|a+be^{x}|=ln|e^{x}|-ln|a+be^{x}|
=ln|(e^{x})/(a+be^{x})|=-ln|(a+be^{x})/(e^{x})|=-ln|ae^{-x}+b|.
Therefore, int 1/(a+be^(x))dx=-(ln|ae^{-x}+b|)/a+C=x/a-(ln|a+be^{x}|)/a+C.