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Câu hỏi: Họ nguyên hàm của hàm số $y=\sqrt{5-3x}$ là:
A. $\dfrac{2}{9}\sqrt{{{\left( 5-3x \right)}^{3}}}+C$
B. $-\dfrac{2}{5}\sqrt{5-3x}+C$
C. $-\dfrac{2}{9}\sqrt{{{\left( 5-3x \right)}^{3}}}+C$
D. $\dfrac{1}{2}\sqrt{5-3x}+C$
A. $\dfrac{2}{9}\sqrt{{{\left( 5-3x \right)}^{3}}}+C$
B. $-\dfrac{2}{5}\sqrt{5-3x}+C$
C. $-\dfrac{2}{9}\sqrt{{{\left( 5-3x \right)}^{3}}}+C$
D. $\dfrac{1}{2}\sqrt{5-3x}+C$
Phương pháp:
Sử dụng công thức: $\int{{{\left( ax+b \right)}^{n}}dx=\dfrac{1}{a}.}\dfrac{{{\left( ax+b \right)}^{n+1}}}{n+1}+C.$
Cách giải:
Ta có: $y=\sqrt{5-3x}={{\left( 5-3x \right)}^{\dfrac{1}{2}}}$
$\Rightarrow \int{\sqrt{5-3x}dx}=\int{{{\left( 5-3x \right)}^{\dfrac{1}{2}}}dx}$
$=-\dfrac{1}{3}.\dfrac{{{\left( 5-3x \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C=-\dfrac{2}{9}\sqrt{{{\left( 5-3x \right)}^{3}}}+C$
Sử dụng công thức: $\int{{{\left( ax+b \right)}^{n}}dx=\dfrac{1}{a}.}\dfrac{{{\left( ax+b \right)}^{n+1}}}{n+1}+C.$
Cách giải:
Ta có: $y=\sqrt{5-3x}={{\left( 5-3x \right)}^{\dfrac{1}{2}}}$
$\Rightarrow \int{\sqrt{5-3x}dx}=\int{{{\left( 5-3x \right)}^{\dfrac{1}{2}}}dx}$
$=-\dfrac{1}{3}.\dfrac{{{\left( 5-3x \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C=-\dfrac{2}{9}\sqrt{{{\left( 5-3x \right)}^{3}}}+C$
Đáp án C.