Solve the initial value problem

xy+(1+1ln(x))y=0\displaystyle {x}{y}'+{\left({1}+\frac{{1}}{{\ln{{\left({x}\right)}}}}\right)}{y}={0} with initial condition y(e)=1\displaystyle {y}{\left({e}\right)}={1}

y(x)=\displaystyle {y}{\left({x}\right)}=