[tex] \huge \pink{ \pmb{ \bf{ \frak{answer}}}}[/tex]

[tex] \sf {Given: - \: \: \: (x + y + 1) \frac{dy}{dx} = 1}[/tex]

[tex] \sf { \frac{dx}{dy} = x + y + + 1}[/tex]

which is a linear differential equation with x as dependent variable:-

[tex] {\bold{ \red{ \star \: { \boxed{Here, \:  P=−1  \: ; Q=y+1}}}}}

[/tex]

[tex] \sf \boxed{Integrating \: \: Factor \:  I.F \:

= {e}^{pdy} }[/tex]

[tex] \bold{I.F. \: = \: {e}^{∫ - dy

}

}

[/tex]

Solution is given by :-

[tex] \sf \: x {e}^{ - y} = ∫

{e}^{ - y} (y + 1)dy

[/tex]

[tex] \sf \: x {e}^{ - y} = ∫

{ye}^{ - y} dy + ∫

{e}^{ - y} dy[/tex]

[tex] \implies \sf \: {xe}^{ - y} = - y {e}^{ - y} + ∫

{e}^{ - y} dy - {e}^{ - y} + k[/tex]

[tex] \implies \sf \: x {e}^{ - y} = - y {e}^{ - y} - {e}^{ - y} - {e}^{ - y} + k[/tex]

[tex] \sf \implies \: {xe}^{ - y} = - {ye}^{ - y} - {2e}^{ - y} + k[/tex]

[tex] \sf \implies \: \: x = {ke}^{y} - (y + 2)[/tex]

[tex] \green{ \pmb{ \bf {\frak{hope \: its \: help \: u}}}}[/tex]

Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given differential equation is

[tex]\rm \: \dfrac{dy}{dx} = x + y + xy + 1 \\ [/tex]

can be rewritten as

[tex]\rm \: \dfrac{dy}{dx} = 1 + x + y + xy \\ [/tex]

can be re-arranged as

[tex]\rm \: \dfrac{dy}{dx} = (1 + x) + (y + xy) \\ [/tex]

[tex]\rm \: \dfrac{dy}{dx} = (1 + x) + y(1 + x) \\ [/tex]

[tex]\rm \: \dfrac{dy}{dx} = (1 + x)(1 + y) \\ [/tex]

On separating the variables, we get

[tex]\rm \: \dfrac{dy}{y + 1} = (1 + x)dx\\ [/tex]

On integrating both sides, we get

[tex]\rm \:\displaystyle\int\rm \dfrac{dy}{y + 1} = \displaystyle\int\rm (1 + x)dx\\ [/tex]

[tex]\rm \: log |y + 1| = \dfrac{ {x}^{2} }{2} + x + c \\ [/tex]

Hence, The general solution of differential equation

[tex]\rm \: \dfrac{dy}{dx} = x + y + xy + 1 \: is \: \\ \: \\ \red{ \rm \: log |y + 1| = \dfrac{ {x}^{2} }{2} + x + c }\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}[/tex]