What is the antiderivative of y/(x-1) if y=x^2-3x+2?

To calculate the antiderivative of the function, we'll
have to determine the indefinite integral of (x^2 - 3x +
2)/(x-1)

We notice that the roots of the numerator are 1
and 2, therefore we can re-write the quadratic as a product of linear
factors:

x^2 - 3x + 2 =
(x-1)(x-2)

We'll calculate the indefinite integral of the
given function:

Int (x^2 - 3x + 2)dx/(x-1) = Int
(x-1)(x-2)dx/(x-1)

We'll simplify and we'll
get:

Int (x-1)(x-2)dx/(x-1) = Int (x-2)dx = Int xdx - 2Int
dx

Int (x^2 - 3x + 2)dx/(x-1) = x^2/2 - 2x +
C

The antiderivative of the function is Int
(x^2 - 3x + 2)dx/(x-1) = x^2/2 - 2x + C.