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Câu hỏi: Cho $I=\int\limits_{0}^{4}{x\sqrt{1+2x }\text{d}x}$ và $u=\sqrt{2x+1}$. Mệnh đề nào dưới đây sai?
A. $I=\dfrac{1}{2}\int\limits_{1}^{3}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\text{d}x}$.
B. $I=\int\limits_{1}^{3}{{{u}^{2}}\left( {{u}^{2}}-1 \right)\text{d}u}$.
C. $I=\dfrac{1}{2}\left. \left( \dfrac{{{u}^{5}}}{5}-\dfrac{{{u}^{3}}}{3} \right) \right|_{1}^{3}$.
D. $I=\dfrac{1}{2}\int\limits_{1}^{3}{{{u}^{2}}\left( {{u}^{2}}-1 \right)\text{d}u}$.
A. $I=\dfrac{1}{2}\int\limits_{1}^{3}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\text{d}x}$.
B. $I=\int\limits_{1}^{3}{{{u}^{2}}\left( {{u}^{2}}-1 \right)\text{d}u}$.
C. $I=\dfrac{1}{2}\left. \left( \dfrac{{{u}^{5}}}{5}-\dfrac{{{u}^{3}}}{3} \right) \right|_{1}^{3}$.
D. $I=\dfrac{1}{2}\int\limits_{1}^{3}{{{u}^{2}}\left( {{u}^{2}}-1 \right)\text{d}u}$.
$I=\int\limits_{0}^{4}{x\sqrt{1+2x}\text{d}x}$
Đặt $u=\sqrt{2x+1}$ $\Rightarrow x=\dfrac{1}{2}\left( {{u}^{2}}-1 \right)$ $\Rightarrow \text{d}x=u \text{d}u$, đổi cận: $x=0\Rightarrow u=1$, $x=4\Rightarrow u=3$.
Khi đó $I=\dfrac{1}{2}\int\limits_{1}^{3}{\left( {{u}^{2}}-1 \right){{u}^{2}}\text{d}u}$.
Đặt $u=\sqrt{2x+1}$ $\Rightarrow x=\dfrac{1}{2}\left( {{u}^{2}}-1 \right)$ $\Rightarrow \text{d}x=u \text{d}u$, đổi cận: $x=0\Rightarrow u=1$, $x=4\Rightarrow u=3$.
Khi đó $I=\dfrac{1}{2}\int\limits_{1}^{3}{\left( {{u}^{2}}-1 \right){{u}^{2}}\text{d}u}$.
Đáp án B.