Find the solution to the following differential equation. Take the constants of integration to be zeros

(1+x)y′+2y=sin⁡(x)1+x\displaystyle {\left({1}+{x}\right)}{y}'+{2}{y}=\frac{{\sin{{\left({x}\right)}}}}{{{1}+{x}}}

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

License