Given the function y=(2x-3)/(x-1)*(x-2). What is the antiderivative of the function [y-1/(x-2)]^2010?

First, we'll try to decompose the function y into partial
fractions.

(2x-3)/(x-1)*(x-2) = A/(x-1) +
B/(x-2)

2x - 3 = x(A+B) - 2A -
B

Comparing both sides, we'll
get:

A+B = 2 => A = 2 - B
(1)

-2A-B = -3 => -2(2 - B) - B =
-3

We'll remove the
brackets:

-4 + 2B - B = -3

B =
4 - 3 => B = 1 => A = 2-1 = 1

y =
(2x-3)/(x-1)*(x-2) = 1/(x-1) + 1/(x-2)

Now, we'll calculate
the difference:

y - 1/(x-2) = 1/(x-1) + 1/(x-2) -
1/(x-2)

We'll eliminate like
terms:

y - 1/(x-2) =
1/(x-1)

We'll raise both sides to 2010
power;

[y - 1/(x-2)]^2010 =
1/(x-1)^2010

We'll calculate the antiderivative of the
function [y - 1/(x-2)]^2010, integrating both sides:

Int [y
- 1/(x-2)]^2010 dx = Int dx/(x-1)^2010

We'll replace x - 1
by t:

x-1 = t

We'll
differentiate both sides:

dx =
dt

Int dx/(x-1)^2010 = Int
dt/t^2010

We'll apply negative power
rule:

Int dt/t^2010  =Int
t^(-2010)dt

Int t^(-2010)dt =
t^(-2010+1)/(-2010+1)

Int t^(-2010)dt = -1/2009*t^2009 +
C

The requested antiderivative of the
function [y - 1/(x-2)]^2010 is F(x) = -1/2009*(x-1)^2009 +
C.