Answer
$y= [c_1+\ln (1+e^{-x} )] e^x +[c_2-e^{-x} +\ln (1+e^{-x} )] e^{2x}$
Work Step by Step
Given: $y''-3y'+2y=\dfrac{1}{1+e^{-x}}$
The particular solution is:
$y_p=u_1 e^x+u_2 e^{2x}$
$u'_1=\dfrac{-e^{-x}}{1+e^{-x}} \implies u_1=\int \dfrac{-e^{-x}}{1+e^{-x}}=\ln (1+e^{-x})$
and $u'_2=\dfrac{e^{x}}{e^{3x}+e^{2x}} \implies u_2=\int \dfrac{e^{x}}{e^{3x}+e^{2x}}=\ln (\dfrac{e^x+1}{e^x}-e^{-x}=\ln (1+e^{-x})-e^{x} $
Thus, $y_p=e^x \ln (1+e^{-x}) +e^{2x} [\ln (1+e^{-x})-e^{x}] $
Hence, $y= [c_1+\ln (1+e^{-x} )] e^x +[c_2-e^{-x} +\ln (1+e^{-x} )] e^{2x}$
