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Câu hỏi: Xét tích phân $I=\int_{0}^{1}{{{e}^{\sqrt{2x+1}}}\text{d}x}$, nếu đặt $u=\sqrt{2x+1}$ thì $I$ bằng
A. $\dfrac{1}{2}\int_{1}^{2}{u{{e}^{u}}\text{du}}$.
B. $\int_{0}^{4}{u{{e}^{u}}\text{du}}$.
C. $\int_{1}^{3}{u{{e}^{u}}\text{du}}$.
D. $\dfrac{1}{2}\int_{1}^{3}{{{e}^{u}}\text{du}}$
A. $\dfrac{1}{2}\int_{1}^{2}{u{{e}^{u}}\text{du}}$.
B. $\int_{0}^{4}{u{{e}^{u}}\text{du}}$.
C. $\int_{1}^{3}{u{{e}^{u}}\text{du}}$.
D. $\dfrac{1}{2}\int_{1}^{3}{{{e}^{u}}\text{du}}$
Đặt $u=\sqrt{2x+1}\Rightarrow {{u}^{2}}=2x+1\Rightarrow u\text{d}u=\text{d}x$.
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow u=1 \\
& x=4\Rightarrow u=3 \\
\end{aligned} \right.$.
Khi đó $I=\int_{1}^{3}{{{e}^{u}}u\text{du}}$.
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow u=1 \\
& x=4\Rightarrow u=3 \\
\end{aligned} \right.$.
Khi đó $I=\int_{1}^{3}{{{e}^{u}}u\text{du}}$.
Đáp án C.