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Câu hỏi: Nguyên hàm của hàm số $f\left( x \right)=\sqrt{3x+2}$ là
A. $\dfrac{3}{2}\dfrac{1}{\sqrt{3x+2}}+C$.
B. $\dfrac{2}{3}(3x+2)\sqrt{3x+2}+C$.
C. $\dfrac{1}{3}(3x+2)\sqrt{3x+2}+C$.
D. $\dfrac{2}{9}(3x+2)\sqrt{3x+2}+C$.
A. $\dfrac{3}{2}\dfrac{1}{\sqrt{3x+2}}+C$.
B. $\dfrac{2}{3}(3x+2)\sqrt{3x+2}+C$.
C. $\dfrac{1}{3}(3x+2)\sqrt{3x+2}+C$.
D. $\dfrac{2}{9}(3x+2)\sqrt{3x+2}+C$.
Cách 1:
+ Đặt: $t=\sqrt{3x+2}\to {{t}^{2}}=3x+2$ $\to \dfrac{2t\text{dt}}{3}=\text{d}x$.
+ Khi đó: $\int{\left( \sqrt{3x+2} \right)}\text{d}x=\int{t.}\dfrac{\text{2tdt}}{3}=\dfrac{2}{3}\int{{{t}^{2}}\text{dt}}=\dfrac{2}{9}{{t}^{3}}+C$.
Vậy $\int{\left( \sqrt{3x+2} \right)}\text{d}x=\dfrac{2}{9}\left( 3x+2 \right)\sqrt{3x+2}+C$.
Cách 2:
+ $\int{\left( \sqrt{3x+2} \right)}\text{d}x={{\int{\left( 3x+2 \right)}}^{\dfrac{1}{2}}}\text{d}x=\dfrac{2}{9}{{\left( 3x+2 \right)}^{\dfrac{3}{2}}}+C=\dfrac{2}{9}\left( 3x+2 \right)\sqrt{3x+2}+C$.
+ Đặt: $t=\sqrt{3x+2}\to {{t}^{2}}=3x+2$ $\to \dfrac{2t\text{dt}}{3}=\text{d}x$.
+ Khi đó: $\int{\left( \sqrt{3x+2} \right)}\text{d}x=\int{t.}\dfrac{\text{2tdt}}{3}=\dfrac{2}{3}\int{{{t}^{2}}\text{dt}}=\dfrac{2}{9}{{t}^{3}}+C$.
Vậy $\int{\left( \sqrt{3x+2} \right)}\text{d}x=\dfrac{2}{9}\left( 3x+2 \right)\sqrt{3x+2}+C$.
Cách 2:
+ $\int{\left( \sqrt{3x+2} \right)}\text{d}x={{\int{\left( 3x+2 \right)}}^{\dfrac{1}{2}}}\text{d}x=\dfrac{2}{9}{{\left( 3x+2 \right)}^{\dfrac{3}{2}}}+C=\dfrac{2}{9}\left( 3x+2 \right)\sqrt{3x+2}+C$.
Đáp án D.