Solve [2x-ln(y+1)] dx - (x+y)/(y+1) dy = 0

Solve $[2x-\ln(y+1)]\,dx - \dfrac{x+y}{y+1}\,dy = 0$.

Check exactness

\(M=2x-\\ln(y+1)\), \(N=-\\dfrac{x+y}{y+1}\). Then \(M_y = -\\dfrac{1}{y+1}\) and \(N_x=-\\dfrac{1}{y+1}\): equal, so the equation is exact. (Confirmed symbolically with sympy.)

Integrate \(M\) with respect to \(x\)

\(F(x,y) = \\int [2x-\\ln(y+1)]\\,dx = x^2 - x\\ln(y+1) + g(y)\).

Determine \(g(y)\)

\(F_y = -\\dfrac{x}{y+1} + g'(y)\) must equal \(N=-\\dfrac{x+y}{y+1}\), so \(g'(y) = -\\dfrac{y}{y+1} = -1+\\dfrac{1}{y+1}\), giving \(g(y) = -y+\\ln(y+1)\) (absorbing the constant of integration into \(C\)).

Assemble the solution

\[ F(x,y) = x^2 - x\\ln(y+1) - y + \\ln(y+1) = x^2 + (1-x)\\ln(y+1) - y = C. \]

Verify

Direct computation confirms \(\\partial F/\\partial x = 2x-\\ln(y+1) = M\) and \(\\partial F/\\partial y = -\\dfrac{x}{y+1}-1+\\dfrac{1}{y+1} = \\dfrac{-x-y}{y+1}=N\) exactly (sympy: both differences simplify to \(0\)). Implicit-differentiating \(F(x,y)=C\) with respect to \(x\) and solving for \(dy/dx\) with sympy gives \(\\dfrac{dy}{dx} = \\dfrac{(2x-\\ln(y+1))(y+1)}{x+y}\), which matches \(-M/N\) from the original equation identically (the symbolic difference simplifies to \(0\)) — no sign ambiguity remains.