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Câu hỏi: Cho $I=\int\limits_{0}^{1}{x{{\left( {{x}^{2}}+1 \right)}^{3}}}\text{d}x$. Nếu đặt $u={{x}^{2}}+1$ thì $I$ bằng
A. $\int\limits_{0}^{1}{{{u}^{3}}}\text{d}u$.
B. $\dfrac{1}{2}\int\limits_{0}^{1}{{{u}^{3}}}\text{d}u$.
C. $\dfrac{1}{2}\int\limits_{1}^{2}{{{u}^{3}}}\text{d}u$.
D. $\int\limits_{1}^{2}{{{u}^{3}}}\text{d}u$.
A. $\int\limits_{0}^{1}{{{u}^{3}}}\text{d}u$.
B. $\dfrac{1}{2}\int\limits_{0}^{1}{{{u}^{3}}}\text{d}u$.
C. $\dfrac{1}{2}\int\limits_{1}^{2}{{{u}^{3}}}\text{d}u$.
D. $\int\limits_{1}^{2}{{{u}^{3}}}\text{d}u$.
Đặt $u={{x}^{2}}+1\Rightarrow \text{d}u=2x\text{d}x\Rightarrow x\text{d}x=\dfrac{1}{2}\text{d}u$
Đổi cận: $\begin{aligned}
& x=0\Rightarrow u=1 \\
& x=1\Rightarrow u=2 \\
\end{aligned}$
Khi đó
$I=\int\limits_{1}^{2}{\dfrac{1}{2}{{u}^{3}}}\text{d}u=\dfrac{1}{2}\int\limits_{1}^{2}{{{u}^{3}}}\text{d}u$.
Đổi cận: $\begin{aligned}
& x=0\Rightarrow u=1 \\
& x=1\Rightarrow u=2 \\
\end{aligned}$
Khi đó
$I=\int\limits_{1}^{2}{\dfrac{1}{2}{{u}^{3}}}\text{d}u=\dfrac{1}{2}\int\limits_{1}^{2}{{{u}^{3}}}\text{d}u$.
Đáp án C.