1. **Problem:** Find the general solution of the differential equation $$(x^2 + 1)dx + xye^y dy = 0.$$ 2. **Step 1: Identify the type of differential equation.** This is a first-order differential equation in the form $$M(x,y)dx + N(x,y)dy = 0$$ where $$M = x^2 + 1$$ and $$N = xye^y.$$ 3. **Step 2: Check if the equation is exact.** Calculate $$\frac{\partial M}{\partial y} = 0$$ and $$\frac{\partial N}{\partial x} = ye^y.$$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact. 4. **Step 3: Try to find an integrating factor.** Check if an integrating factor depends on $$x$$ or $$y$$ alone. Calculate $$\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{0 - ye^y}{xye^y} = -\frac{1}{x}.$$ This depends only on $$x$$, so an integrating factor $$\mu(x)$$ exists. 5. **Step 4: Find the integrating factor.** $$\mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}.$$ 6. **Step 5: Multiply the entire equation by $$\mu(x) = \frac{1}{x}$$:** $$\frac{x^2 + 1}{x} dx + ye^y dy = 0$$ which simplifies to $$\left(x + \frac{1}{x}\right) dx + ye^y dy = 0.$$ 7. **Step 6: Check if the new equation is exact.** Let $$\tilde{M} = x + \frac{1}{x}$$ and $$\tilde{N} = ye^y.$$ Calculate $$\frac{\partial \tilde{M}}{\partial y} = 0$$ and $$\frac{\partial \tilde{N}}{\partial x} = 0.$$ Since both partial derivatives are equal, the equation is exact. 8. **Step 7: Find the potential function $$\Psi(x,y)$$ such that:** $$\frac{\partial \Psi}{\partial x} = \tilde{M} = x + \frac{1}{x}$$ Integrate with respect to $$x$$: $$\Psi = \int \left(x + \frac{1}{x}\right) dx = \frac{x^2}{2} + \ln|x| + h(y),$$ where $$h(y)$$ is a function of $$y$$. 9. **Step 8: Differentiate $$\Psi$$ with respect to $$y$$ and set equal to $$\tilde{N}$$:** $$\frac{\partial \Psi}{\partial y} = h'(y) = ye^y.$$ Integrate with respect to $$y$$: $$h(y) = \int ye^y dy.$$ Use integration by parts: Let $$u = y$$, $$dv = e^y dy$$, then $$du = dy$$, $$v = e^y$$. So, $$h(y) = y e^y - \int e^y dy = y e^y - e^y + C = e^y(y - 1) + C.$$ 10. **Step 9: Write the implicit solution:** $$\Psi(x,y) = \frac{x^2}{2} + \ln|x| + e^y(y - 1) = C,$$ where $$C$$ is an arbitrary constant. **Final answer:** $$\boxed{\frac{x^2}{2} + \ln|x| + e^y(y - 1) = C}.$$