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Câu hỏi: Với cách biến đổi $u=\sqrt{4x+5}$ thì tích phân $\int\limits_{-1}^{1}{x\sqrt{4x+5}dx}$ trở thành
A. $\int\limits_{-1}^{1}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}du}$.
B. $\int\limits_{1}^{3}{\dfrac{u\left( {{u}^{2}}-5 \right)}{8}du}$.
C. $\int\limits_{1}^{3}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{4}du}$.
D. $\int\limits_{1}^{3}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}du}$.
A. $\int\limits_{-1}^{1}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}du}$.
B. $\int\limits_{1}^{3}{\dfrac{u\left( {{u}^{2}}-5 \right)}{8}du}$.
C. $\int\limits_{1}^{3}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{4}du}$.
D. $\int\limits_{1}^{3}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}du}$.
Đặt $u=\sqrt{4x+5}\Leftrightarrow {{u}^{2}}=4x+5\Rightarrow \left\{ \begin{aligned}
& x=\dfrac{{{u}^{2}}-5}{4} \\
& 2udu=4dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=\dfrac{{{u}^{2}}-5}{4} \\
& dx=\dfrac{udu}{2} \\
\end{aligned} \right.$.
Đổi cận: $x=-1\Rightarrow u=1,x=1\Rightarrow u=3$
Khi đó $\int\limits_{-1}^{1}{x\sqrt{4x+5}dx}=\int\limits_{1}^{3}{\dfrac{{{u}^{2}}-5}{4}.u.\dfrac{udu}{2}=\int\limits_{1}^{3}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}du}}$.
& x=\dfrac{{{u}^{2}}-5}{4} \\
& 2udu=4dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=\dfrac{{{u}^{2}}-5}{4} \\
& dx=\dfrac{udu}{2} \\
\end{aligned} \right.$.
Đổi cận: $x=-1\Rightarrow u=1,x=1\Rightarrow u=3$
Khi đó $\int\limits_{-1}^{1}{x\sqrt{4x+5}dx}=\int\limits_{1}^{3}{\dfrac{{{u}^{2}}-5}{4}.u.\dfrac{udu}{2}=\int\limits_{1}^{3}{\dfrac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}du}}$.
Đáp án D.