Giải thích các bước giải:
$y'=\dfrac{\dfrac{1}{x}(x+1)-\ln x}{(x+1)^2}=\dfrac{x+1-x\ln x}{x(x+1)^2}$
$y''=(\dfrac{x+1-x\ln x}{x(x+1)^2})'=\dfrac{2x^3\ln x-3x^3+2x^2\ln x-7x^2-5x-1}{x^2(x+1)^4}$
$\to 2y'+(x+1)y''=2.\dfrac{x+1-x\ln x}{x(x+1)^2}+(x+1).\dfrac{2x^3\ln x-3x^3+2x^2\ln x-7x^2-5x-1}{x^2(x+1)^4}=\dfrac{-1}{x^2}$
$\to 2y'+(x+1)y''+\dfrac{1}{x^2}=0\to A$
