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(C) Let $I = \int \frac{x+1}{(x-2) \sqrt{1-x}} d x$. Put $1-x = t^2$,so $x = 1-t^2$ and $dx = -2t dt$. Substituting these into the integral: $I = \int

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$6 \operatorname{Tan}^{-1} \sqrt{1-x} - 2 \sqrt{1-x} + c$

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(C) Let $I = \int \frac{x+1}{(x-2) \sqrt{1-x}} d x$. Put $1-x = t^2$,so $x = 1-t^2$ and $dx = -2t dt$. Substituting these into the integral: $I = \int \frac{(1-t^2)+1}{(1-t^2-2) \cdot t} (-2t) dt$ $I = \int \frac{2-t^2}{(-1-t^2) \cdot t} (-2t) dt$ $I = \int \frac{2-t^2}{-(1+t^2)} (-2) dt = 2 \int \frac{2-t^2}{1+t^2} dt$ $I = 2 \int \frac{3-(1+t^2)}{1+t^2} dt = 2 \int \left( \frac{3}{1+t^2} - 1 \right) dt$ $I = 6 \int \frac{1}{1+t^2} dt - 2 \int 1 dt$ $I = 6 \operatorname{Tan}^{-1}(t) - 2t + c$ Substituting $t = \sqrt{1-x}$ back: $I = 6 \operatorname{Tan}^{-1}(\sqrt{1-x}) - 2\sqrt{1-x} + c$.

$\int \frac{x+1}{(x-2) \sqrt{1-x}} d x=$

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