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

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The function y = (2x-3)/(x-1)*(x-2)

We have to find the antiderivative of [y-1/(x-2)]^2010

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

=> [(2x-3)/(x-1)*(x-2) - 1/(x - 2)]^2010

=> [(2x-3)/(x-1)*(x-2) - (x - 1)/(x - 1)(x - 2)]^2010

=> [((2x-3) - (x - 1))/(x-1)*(x-2)]^2010

=> [(2x - 3 - x + 1)/(x-1)*(x-2)]^2010

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

=> [1/(x-1)]^2010

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

let x - 1 = y, dy = dx

=> Int[1/y^2010 dy]

=> Int[y^-2010 dy]

=> y^-2009/(-2009)

substitute y = x - 1

=> -1/2009*(x - 1)^2009 + C

**The required antiderivative is -1/2009*(x - 1)^2009 + C**

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.**