(x-1)y' + y = x^2 -1 Solve the first-order differential equation

Given (x-1)y'+y=x^2-1
when the first order linear ordinary differential equation has the form of
y'+p(x)y=q(x)
then the general solution is,
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)
so,
(x-1)y'+y=x^2-1
=> (x-1)[y' + y/(x-1)] = x^2 -1
=> y'+y/(x-1)= ((x+1)(x-1))/(x-1)
=> y'+y/(x-1)= (x+1) --------(1)
 
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = 1/(x-1) and q(x)=(x+1)
so on solving with the above general solution we get:
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)
=((int e^(int (1/(x-1))dx) *(x+1)) dx +c)/e^(int(1/(x-1)) dx)
first we shall solve
e^(int (1/(x-1)) dx)=e^(ln|x-1|) = |x-1|  
When x-1<=0 then ln(x-1) is undefined , so  
e^(int(1/(x-1)) dx)=x-1
so
proceeding further, we get
y(x) =((int e^(int (1/(x-1))dx) *(x+1)) dx +c)/e^(int(1/(x-1)) dx)
=((int (x-1)*(x+1)) dx +c)/(x-1)
=((int (x^2-1) ) dx +c)/(x-1)
= (x^3/3 -x  +c)/(x-1)
 
y(x)=(x^3/3 -x +c)/(x-1)