Toán tính nguyên hàm : căn bậc 2( (2/x) -1 ) 12/09/2021 By Ariana tính nguyên hàm : căn bậc 2( (2/x) -1 )
Đáp án: $\sqrt{\dfrac{2-x}{x}}x+\arcsin \left(x-1\right)+C$ Giải thích các bước giải: $\int\sqrt{\dfrac2x-1}dx$ $=\sqrt{\dfrac{2-x}{x}}x-\int \:-\dfrac{1}{x^{\dfrac{1}{2}}\sqrt{-x+2}}dx$ $=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{-x^2+2x}}dx$ $=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{-x^2+2x-1+1}}dx$ $=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{-\left(x-1\right)^2+1}}dx$ $=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{1-\left(x-1\right)^2}}dx$ $=\sqrt{\dfrac{2-x}{x}}x+\arcsin \left(x-1\right)+C$ Trả lời
Đáp án: $\sqrt{\dfrac{2-x}{x}}x+\arcsin \left(x-1\right)+C$
Giải thích các bước giải:
$\int\sqrt{\dfrac2x-1}dx$
$=\sqrt{\dfrac{2-x}{x}}x-\int \:-\dfrac{1}{x^{\dfrac{1}{2}}\sqrt{-x+2}}dx$
$=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{-x^2+2x}}dx$
$=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{-x^2+2x-1+1}}dx$
$=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{-\left(x-1\right)^2+1}}dx$
$=\sqrt{\dfrac{2-x}{x}}x+\int \dfrac{1}{\sqrt{1-\left(x-1\right)^2}}dx$
$=\sqrt{\dfrac{2-x}{x}}x+\arcsin \left(x-1\right)+C$