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Câu hỏi: Cho $\int{2x\left( 3x-2 \right)dx}=A{{\left( 3x-2 \right)}^{8}}+B{{\left( 3x-2 \right)}^{7}}+C$ với $A,B,C\in \mathbb{R}$. Tính giá trị của biểu thức $12A+7B$.
A. $\dfrac{23}{252}.$
B. $\dfrac{241}{252}.$
C. $\dfrac{52}{9}.$
D. $\dfrac{7}{9}.$
A. $\dfrac{23}{252}.$
B. $\dfrac{241}{252}.$
C. $\dfrac{52}{9}.$
D. $\dfrac{7}{9}.$
Ta có: $I=\int{2x{{\left( 3x-2 \right)}^{6}}dx}$.
Đặt $3x-2=t\Rightarrow x=\dfrac{t+2}{3}\Rightarrow dx=\dfrac{1}{3}dt$.
Suy ra $I=\int{\dfrac{2}{9}\left( t+2 \right){{t}^{6}}dt}=\dfrac{2}{9}\int{\left( {{t}^{7}}+2{{t}^{6}} \right)dt}=\dfrac{2}{9}\left( \dfrac{{{t}^{8}}}{8}+\dfrac{2{{t}^{7}}}{7} \right)+C=\dfrac{1}{36}{{t}^{8}}+\dfrac{4}{63}{{t}^{7}}+C$
$\Rightarrow I=\dfrac{1}{36}{{\left( 3x-2 \right)}^{8}}+\dfrac{4}{63}{{\left( 3x-2 \right)}^{7}}+C\Rightarrow \left\{ \begin{aligned}
& A=\dfrac{1}{36} \\
& B=\dfrac{4}{63} \\
\end{aligned} \right.\Rightarrow 12A+7B=12.\dfrac{1}{36}+7.\dfrac{4}{63}=\dfrac{7}{9}.$
Đặt $3x-2=t\Rightarrow x=\dfrac{t+2}{3}\Rightarrow dx=\dfrac{1}{3}dt$.
Suy ra $I=\int{\dfrac{2}{9}\left( t+2 \right){{t}^{6}}dt}=\dfrac{2}{9}\int{\left( {{t}^{7}}+2{{t}^{6}} \right)dt}=\dfrac{2}{9}\left( \dfrac{{{t}^{8}}}{8}+\dfrac{2{{t}^{7}}}{7} \right)+C=\dfrac{1}{36}{{t}^{8}}+\dfrac{4}{63}{{t}^{7}}+C$
$\Rightarrow I=\dfrac{1}{36}{{\left( 3x-2 \right)}^{8}}+\dfrac{4}{63}{{\left( 3x-2 \right)}^{7}}+C\Rightarrow \left\{ \begin{aligned}
& A=\dfrac{1}{36} \\
& B=\dfrac{4}{63} \\
\end{aligned} \right.\Rightarrow 12A+7B=12.\dfrac{1}{36}+7.\dfrac{4}{63}=\dfrac{7}{9}.$
Đáp án D.