Verified answer

∫ 1 / (x-2)^2 - 1 dx

∫ 1 / (x-2)^2 dx - ∫ 1 dx

∫ 1 / (x-2)^2 dx - x

For the first part of the integral, let,

u = x-2 dx

du = dx

∫ 1/u^2 du - x

∫ u^-2 du - x

-u^-1 - x + C

-(1/(x-2)) - x + C

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