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Ta có
$\dfrac{1}{(x-1)\sqrt{x^2-4x+3}}=\dfrac{x-3}{(x-1)(x-3)\sqrt{x^2-4x+3}}$
$\dfrac{1}{(x-1)\sqrt{x^2-4x+3}}=\dfrac{\dfrac{(x^2-4x+3)-(x-2)(x-3)}{\sqrt{x^2-4x+3}}}{x^2-4x+3}$
$\dfrac{1}{(x-1)\sqrt{x^2-4x+3}}=\dfrac{\sqrt{x^2-4x+3}-\dfrac{(x-2)(x-3)}{\sqrt{x^2-4x+3}}}{x^2-4x+3}$
$\dfrac{1}{(x-1)\sqrt{x^2-4x+3}}=\dfrac{\sqrt{x^2-4x+3}(x-3)'-(\sqrt{x^2-4x+3})'(x-3)}{x^2-4x+3}$
$\dfrac{1}{(x-1)\sqrt{x^2-4x+3}}=\left ( \dfrac{x-3}{\sqrt{x^2-4x+3}} \right )'$
Suy ra
$\int\limits_{2+\sqrt{2}}^{4}\dfrac{1}{(x-1)\sqrt{x^{2}-4x+3}}dx=\left[ {\dfrac{x-3}{\sqrt{x^2-4x+3}}} \right]_{2+\sqrt{2}}^{4}=1-\sqrt 2+\dfrac{1}{\sqrt 3}$
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