Nguyên hàm của hàm số $\Large\mathrm{y=\sqrt{\dfrac{x+1}{x-2}} . \dfra

Nguyên hàm của hàm số $\Large\mathrm{y=\sqrt{\dfrac{x+1}{x-2}} . \dfrac{1}{(x-2)^{2}}}$ bằng

• A.

  $\Large\mathrm{\dfrac{2}{9} \sqrt{\left(\dfrac{x+1}{x-2}\right)^{3}}+C}$

• B.

  $\Large\mathrm{\dfrac{2}{3} \sqrt{\left(\dfrac{x+1}{x-2}\right)^{3}}+C}$

• C.

  $\Large\mathrm{\sqrt{\left(\dfrac{x+1}{x-2}\right)^{3}}+C}$

• D.

  $\Large\mathrm{-\dfrac{2}{9} \sqrt{\left(\dfrac{x+1}{x-2}\right)^{3}}+C}$

Tính $\Large\mathrm{I=\int \sqrt{\dfrac{x+1}{x-2}} . \dfrac{1}{(x-2)^{2}} d x}$. Đặt $\Large\mathrm{t=\sqrt{\dfrac{x+1}{x-2}} \Rightarrow t^{2}=\dfrac{x+1}{x-2} \Rightarrow 2 t d t=\dfrac{-3}{(x-2)^{2}} d x}$$\Large  \Leftrightarrow \dfrac{d x}{(x-2)^{2}}=-\dfrac{2}{3} t d t$
Khi đó: $\Large I=\int t .\left(-\dfrac{2}{3} t d t\right)=-\dfrac{2}{3} \int t^{2} d t=-\dfrac{2}{9} t^{3}+C$ $\Large =-\dfrac{2}{9} \sqrt{\left(\dfrac{x+1}{x-2}\right)^{3}}+C \Rightarrow$ đáp án D.