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Câu hỏi: Nguyên hàm của hàm só $f\left( x \right)=\sqrt{2\text{x}-1}$ là
A. $\int{f\left( x \right)d\text{x}=\dfrac{2}{3}\left( 2\text{x}-1 \right)\sqrt{2\text{x}-1}+C.}$
B. $\int{f\left( x \right)d\text{x}=\dfrac{1}{3}\left( 2\text{x}-1 \right)\sqrt{2\text{x}-1}+C.}$
C. $\int{f\left( x \right)d\text{x}=-\dfrac{1}{3}\sqrt{2\text{x}-1}+C.}$
D. $\int{f\left( x \right)d\text{x}=\dfrac{1}{2}\sqrt{2\text{x}-1}+C.}$
A. $\int{f\left( x \right)d\text{x}=\dfrac{2}{3}\left( 2\text{x}-1 \right)\sqrt{2\text{x}-1}+C.}$
B. $\int{f\left( x \right)d\text{x}=\dfrac{1}{3}\left( 2\text{x}-1 \right)\sqrt{2\text{x}-1}+C.}$
C. $\int{f\left( x \right)d\text{x}=-\dfrac{1}{3}\sqrt{2\text{x}-1}+C.}$
D. $\int{f\left( x \right)d\text{x}=\dfrac{1}{2}\sqrt{2\text{x}-1}+C.}$
Ta có: $\int{f\left( x \right)dx=\int{\sqrt{2x-1}dx=\int{{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}dx=\dfrac{1}{2}.\dfrac{{{\left( 2x-1 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C}}}$
$=\dfrac{1}{2}.\dfrac{3}{2}.\sqrt{{{\left( 2x-1 \right)}^{3}}}+C=\dfrac{1}{3}.\left( 2x-1 \right).\sqrt{2x-1}+C.$
$=\dfrac{1}{2}.\dfrac{3}{2}.\sqrt{{{\left( 2x-1 \right)}^{3}}}+C=\dfrac{1}{3}.\left( 2x-1 \right).\sqrt{2x-1}+C.$
Đáp án B.