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The question is to find the reduction formula of $\ I_{n}=\int \left ( \frac{x-1}{x-2} \right )^{n} (x) dx $

I was asked to provide a recurrent form of this integral involving $\ I_{n-1}$ I've tried with the substitution: Let $\ u = x -2 $, $\ du = dx $

Then $\ I_{n}=\int \left ( \frac{u+1}{u} \right )^{n} (u+2) du $

Tried by expanding this one step: $\ I_{n}=\int \left ( \frac{u+1}{u} \right )^{n-1} ( \frac{u+1}{u})(u+2) du $

Further , I tried integrating by parts, without luck.

This is my first semester taking Calculus so I've tried, but after spending three days trying to solve this I'm lost, if someone can try I'd appreciate it very much.

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$$I_{n+3} - 3I_{n+2} + 3I_{n+1} - I_n = \int \left( \frac{x-1}{x-2} \right)^n \frac{x}{x-2} \frac{1}{(x-2)^2}\, \text{d}x$$

The RHS has a completely closed form which can be found by using IBP twice.

This reduction formula was found entirely through trial and error.

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