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Câu hỏi: Thực hiện phép biến đổi $t=\sqrt[3]{3x+1}$ thì tích phân $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{x+1}{\sqrt[3]{3x+1}}.}\text{d}x=\int\limits_{1}^{2}{g\left( t \right)\text{.d}t}$. Khi đó:
A. $g\left( 3 \right)=31.$
B. $g\left( 3 \right)=29.$
C. $g\left( 3 \right)=33.$
D. $g\left( 3 \right)=25.$
A. $g\left( 3 \right)=31.$
B. $g\left( 3 \right)=29.$
C. $g\left( 3 \right)=33.$
D. $g\left( 3 \right)=25.$
Đặt $t=\sqrt[3]{3x+1}\Rightarrow {{t}^{3}}=3x+1\Rightarrow x=\dfrac{{{t}^{3}}-1}{3}$ và $dx={{t}^{2}}dt$. Đổi cận: $x=0\Rightarrow t=1; x=\dfrac{7}{3}\Rightarrow t=2.$
Khi đó $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{x+1}{\sqrt[3]{3x+1}}.}dx=\int\limits_{1}^{2}{\dfrac{\dfrac{{{t}^{3}}-1}{3}+1}{t}.{{t}^{2}}dt}=\int\limits_{1}^{2}{\dfrac{{{t}^{3}}+2}{3t}.{{t}^{2}}dt}=\int\limits_{1}^{2}{\dfrac{{{t}^{4}}+2t}{3}dt=}\int\limits_{1}^{2}{g\left( t \right).dt}$.
Suy ra $g\left( t \right)=\dfrac{{{t}^{4}}+2t}{3}\Rightarrow g\left( 3 \right)=29$.
Khi đó $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{x+1}{\sqrt[3]{3x+1}}.}dx=\int\limits_{1}^{2}{\dfrac{\dfrac{{{t}^{3}}-1}{3}+1}{t}.{{t}^{2}}dt}=\int\limits_{1}^{2}{\dfrac{{{t}^{3}}+2}{3t}.{{t}^{2}}dt}=\int\limits_{1}^{2}{\dfrac{{{t}^{4}}+2t}{3}dt=}\int\limits_{1}^{2}{g\left( t \right).dt}$.
Suy ra $g\left( t \right)=\dfrac{{{t}^{4}}+2t}{3}\Rightarrow g\left( 3 \right)=29$.
Đáp án B.