First you'll need the complementary solutions. The characteristic equation for this ODE is
\(D^2+2D+1=(D+1)^2=0\implies D=-1\)
with multiplicity 2. This means two linearly independent solutions will be \(y_1=e^{-x}\) and \(y_2=xe^{-x}\).
Via variation of parameters, the particular solution will take the form
\(y_p=y_1u_1+y_2u_2\)
where
\(u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx\)
\(u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx\)
The Wronskian is
\(W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}\)
So, you have
\(u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx\)
\(u_1=-\displaystyle\int x\log x\,\mathrm dx\)
\(u_1=\dfrac14x^2-\dfrac12x^2\log x\)
and
\(u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx\)
\(u_2=\displaystyle\int\log x\,\mathrm dx\)
\(u_2=x(\log x-1)\)
So the solution to the ODE is
\(y=C_1y_1+C_2y_2+y_1u_1+y_2u_2\)
\(y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}\)
\(y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)\)
