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Câu hỏi: Xét $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{(x+1)\text{d}x}{\sqrt[3]{3x+1}}}$, nếu đặt $t=\sqrt[3]{3x+1}$ $t=\sqrt[3]{3x+1}$ thì $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{(x+1)\text{d}x}{\sqrt[3]{3x+1}}}$ bằng
A. $\dfrac{1}{3}\int\limits_{1}^{2}{({{t}^{4}}+2t)\text{d}t.}$
B. $\dfrac{1}{3}\int\limits_{1}^{4}{({{t}^{4}}-2t)\text{d}t.}$
C. $3\int\limits_{1}^{2}{({{t}^{4}}+2t)\text{d}t.}$
D. $\dfrac{1}{3}\int\limits_{0}^{2}{({{t}^{4}}+4t)\text{d}t.}$
A. $\dfrac{1}{3}\int\limits_{1}^{2}{({{t}^{4}}+2t)\text{d}t.}$
B. $\dfrac{1}{3}\int\limits_{1}^{4}{({{t}^{4}}-2t)\text{d}t.}$
C. $3\int\limits_{1}^{2}{({{t}^{4}}+2t)\text{d}t.}$
D. $\dfrac{1}{3}\int\limits_{0}^{2}{({{t}^{4}}+4t)\text{d}t.}$
Đặt $t=\sqrt[3]{3x+1}\Rightarrow x=\dfrac{{{t}^{3}}-1}{3}\Rightarrow dx={{t}^{2}}dt$.
Khi đó $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{(x+1)dx}{\sqrt[3]{3x+1}}}=\int\limits_{1}^{2}{\dfrac{\left( \dfrac{{{t}^{3}}-1}{3}+1 \right){{t}^{2}}dt}{t}}=\dfrac{1}{3}\int\limits_{1}^{2}{({{t}^{4}}+2t)dt}$.
Khi đó $\int\limits_{0}^{\dfrac{7}{3}}{\dfrac{(x+1)dx}{\sqrt[3]{3x+1}}}=\int\limits_{1}^{2}{\dfrac{\left( \dfrac{{{t}^{3}}-1}{3}+1 \right){{t}^{2}}dt}{t}}=\dfrac{1}{3}\int\limits_{1}^{2}{({{t}^{4}}+2t)dt}$.
Đáp án A.