Given Differential Equation :
(x + y + 1) \(\frac {dy}{dx}\) = 1
Formula :
i) \(\int\) 1dx = x
ii) \(\int\) u. v dx = u. \(\int\) v. dx - \(\int\) \((\frac{du}{dx}.\int v\, dx)\) dx
iii) \(\int\) ekx dx = \(\frac {e^{kx}}{k}\)
iv) \(\frac{d}{dx}\) (xn) = nxn-1
v) General solution :
For the differential equation in the form of
\(\frac{dx}{dy} \,+ Px\, =Q\)
General solution is given by,
x. (I.F.) = \(\int\) Q. (I.F.) dy + c
Where, integrating factor,
I.F. = \(e^{\int p\, dx}\)
Given differential equation is

Equation (1) is of the form
\(\frac{dx}{dy} \,+ Px\, =Q\)
where, P= - 1 and Q = y + 1
Therefore, integrating factor is

General solution is

Let, u=y+1 and v= e-y

Substituting I in eq(2),

Dividing above equation by e-y

Therefore, general solution is
x cey - (y + 2)