Now separating variable x on one side and variable y on another side, we have

Adding 1 and subtracting 1 to the numerator of LHS, we get

Integrating both sides using identities:

And

Assuming e^x = t and differentiating both sides we get,

e^xdx = dt

substituting this value in above equation

y – log(y+1) = log(1+t)

substituting t as e^x

y – log(y+1) = log(1+e^x) + c