Since it is a form of linear differential equation where

P = g’(x) and Q = g(x) g’(x)

Integrating Factor (I.F) = e^{∫ p dx}

I.F = e^{∫ g’(x) dx} = e^g(x)

Solution of differential equation is given by

y.(I.F) = ∫ Q.(I.F) dx + C

⇒ y. e^g(x) = ∫ g(x). g’(x). e^g(x) dx + C

Consider integral ∫ g(x). g’(x). e^g(x) dx

Put g(x) = t

⇒ g’(x) dx = dt

⇒ ∫ t. e^t dt

Treating t as first function and e^t as second function, So integrating by Parts we get,

⇒ t. e^t - ∫ 1.e^t dt + C

⇒ e^t (t – 1) + C

Putting value of t we get,

⇒ e ^g(x) (g(x) – 1) + C

∴ y. e ^g(x) = e ^g(x) (g(x) – 1) + C

Dividing e ^g(x) both sides we get,

⇒ y = (g(x) – 1) + C e^-g(x)

⇒ y - g(x) + 1 = C e^-g(x)

Taking log both sides we get,

⇒ log (y – g(x) + 1) = log (C e^-g(x))

⇒ log (y – g(x) + 1) = log C – g(x) log e

⇒ log (y – g(x) + 1) = log C – g(x) ∵ log e = 1

⇒ g(x) + log {1 + y – g(x)} = C ⇒ (B) where log C = C