first order differential equation?

Solution of this differential equation involves inverse hyperbolic tangent, which is a solution to the following integral:

∫ 1/(u² - 1) du = artanh(t) = (1/2)∙ln( (1+ u)/(1 - u) )

To solve the differential equation separate variables and integrate:

dx/dt = (x - 1)² - x

<=>

dx/dt = x² - 3∙x + 1

<=>

∫ 1/(x² - 3∙x + 1) dx = ∫ dt

<=>

∫ 1/( x² - 3∙x + (9/4) - (1/4)) dx = ∫ dt

<=>

∫ 1/( (x - (3/2))² - (1/4)) dx = ∫ dt

<=>

∫ 4/( [2∙(x - (3/2))]² - 1) dx = ∫ dt

on left hind side substitute

u = 2∙(x - (3/2)) => du= 2 dx

∫ 2/(u² - 1) dx = ∫ dt

<=>

2∙artanh( u ) = t + c

<=>

artanh( 2∙(x - (3/2)) ) = (1/2)∙t + C

with C = c/2

<=>

x = (3/2) + (1/2)∙tanh( (1/2)∙t + C )

dx/dt = (x-1)^2 - x

dx/((x-1)^2 - x) = dt

t +C =∫ ( 1/(x^2 - 3x +1) dx

Now ∫1/((x-3/2)^2 - (√5/2)^2) dx =

(1/√5)∫1/((x-3/2) - (√5/2)) dx - (1/√5)∫1/((x-3/2) + (√5/2)) dx =

(1/√5)Ln((x-3/2) - (√5/2)) - (1/√5)Ln((x-3/2) + (√5/2)) =

(1/√5)Ln(((x-3/2) - (√5/2))/((x-3/2) + (√5/2)))

e^(√5t+C√5) = ((x-3/2) - (√5/2))/((x-3/2) + (√5/2))

e^(√5t+C√5) /(1- e^(√5t+C√5)) = ((x-3/2) - (√5/2))/√5

x = (3/2)+ √5e^(√5t+C√5) /(1- e^(√5t+C√5)) + (√5/2)