<s>[INST] compute an integral $\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}$ How we can compute this integral:
$$\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}$$
i know that the solution is:
$$I=\frac{2 (x-2)}{\sqrt{x^2-3x+2}}$$
 [/INST] Substitute $\displaystyle t=\frac{\sqrt{x^2-3x+2}}{x-1}$ to get $x=\dfrac{t^2-2}{t^2-1}$ and $dx=\dfrac{2t}{(1-t^2)^2}dt$;  then
$\displaystyle\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}=\int\frac{1}{\frac{1}{(1-t^2)^2}\cdot t}\cdot\frac{2t}{(1-t^2)^2} dt=2\int dt=2t+C$
$\displaystyle=\frac{2\sqrt{x^2-3x+2}}{x-1}+C=\frac{2(x-2)}{\sqrt{x^2-3x+2}}+C$
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