令 z = y' 则 z' = y''So (x + 1)y'' + y' = ln(x + 1)--> (x + 1)z' + z = ln(x + 1)--> (x + 1)z' + z(x + 1) = ln(x + 1)--> [z(x + 1)]' = ln(x + 1)--> z(x + 1) = ∫ ln(x + 1) dx--> z(x + 1) = xln(x + 1) -...