kanorkant
[66280]
06/01/2020 1:46:20 PM
Nguyên hàm chứa căn
$\displaystyle\int{\dfrac{x}{\sqrt{2x-{{x}^{2}}}}dx}=\dfrac{1}{2}\displaystyle\int{\left( \dfrac{2x-2}{\sqrt{2x-{{x}^{2}}}}+\dfrac{2}{\sqrt{2x-{{x}^{2}}}} \right)dx}\\=-\dfrac{1}{2}\displaystyle\int{\dfrac{d\left( -{{x}^{2}}+2x \right)}{\sqrt{2x-{{x}^{2}}}}}+\displaystyle\int{\dfrac{1}{\sqrt{1-{{(x-1)}^{2}}}}dx}=-\sqrt{2x-{{x}^{2}}}+\displaystyle\int{\dfrac{1}{\sqrt{1-{{(x-1)}^{2}}}}dx}.$
Đặt $(x-1)=\sin t\Rightarrow dx=\cos tdt.$
Ta được $\displaystyle\int{\dfrac{1}{\sqrt{1-{{(x-1)}^{2}}}}dx}=\displaystyle\int{\dfrac{1}{\sqrt{1-{{\sin }^{2}}t}}.\cos tdt}=\displaystyle\int{dt}=t+C=\arcsin (x-1)+C.$
Vậy $\displaystyle\int{\dfrac{x}{\sqrt{2x-{{x}^{2}}}}dx}=-\sqrt{2x-{{x}^{2}}}+\arcsin (x-1)+C.$
\[\int {\frac{x}{{\sqrt {2x - {x^2}} }}dx} = \int {\sqrt {\frac{x}{{2 - x}}} dx} ;t = \sqrt {\frac{x}{{2 - x}}} .....\]