separate the differential equation into variables
xydx + (x+1)dy=0
"xydx + (x+1)dy=0\\\\\nxydx =- (x+1)dy\\\\\n\\frac{xdx}{x+1}=-\\frac{dy}{y}"
Let, "x+1=t\\Rightarrow dx=dt"
"\\frac{t-1}{t}dt=-\\frac{dy}{y}"
"(1-\\frac{1}{t})dt=" "-\\frac{dy}{y}"
Integrating both sides, we get
"t-\\ln t=-\\ln y+c"
"x+1-\\ln (x+1)=-\\ln y+c"
"x-\\ln (x+1)+\\ln y+1=c"
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