Tích phân  $\int\limits_{2+\sqrt{2}}^{4}\frac{1}{(x-1)\sqrt{x^{2}-4x+3}}dx$
$\int\limits_{2+\sqrt{2}}^{4}\frac{1}{(x-1)\sqrt{(x-3)(x-1)}}dx=\int\limits_{2+\sqrt{2}}^{4}\frac{1}{(x-1)^{2}\sqrt{\frac{x-3}{x-1}}}dx$
dat t=$\sqrt{\frac{(x-3)}{x-1)}}  => tdt=\frac{1}{(x-1)^{2}}dx=>I=\int\limits_{\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}}^{\frac{1}{\sqrt{3}}}dt=\frac{1}{\sqrt{3}}-\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ 
ngan gon vay thoi –  long_teenboy 16-01-13 10:32 PM
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 

 $\int\limits_{2+\sqrt{2}}^{4}\dfrac{1}{(x-1)\sqrt{x^{2}-4x+3}}dx=\left[ {\dfrac{x-3}{\sqrt{x^2-4x+3}}} \right]_{2+\sqrt{2}}^{4}=1-\sqrt 2+\dfrac{1}{\sqrt 3}$
Hãy ấn chữ V dưới đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Thanks! –  Trần Nhật Tân 16-01-13 02:37 PM

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