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Câu hỏi: Khi tính nguyên hàm $\int{\dfrac{x-2021}{\sqrt{x+1}}\text{d}x}$, bằng cách đặt $u=\sqrt{x+1}$ ta được nguyên hàm nào dưới đây?
A. $2\int{u\left( {{u}^{2}}-2022 \right)\text{d}u}$.
B. $\int{\left( {{u}^{2}}-2022 \right)\text{d}u}$.
C. $2\int{\left( {{u}^{2}}-2022 \right)\text{d}u}$.
D. $2\int{\left( {{u}^{2}}-2021 \right)\text{d}u}$.
A. $2\int{u\left( {{u}^{2}}-2022 \right)\text{d}u}$.
B. $\int{\left( {{u}^{2}}-2022 \right)\text{d}u}$.
C. $2\int{\left( {{u}^{2}}-2022 \right)\text{d}u}$.
D. $2\int{\left( {{u}^{2}}-2021 \right)\text{d}u}$.
Đặt $u=\sqrt{x+1}$, $u\ge 0$ nên ${{u}^{2}}=x+1$ $\Rightarrow \left\{ \begin{aligned}
& \text{d}x=2u\text{d}u \\
& x={{u}^{2}}-1 \\
\end{aligned} \right.$.
.Khi đó $\int{\dfrac{x-2021}{\sqrt{x+1}}\text{d}x}=\int{\dfrac{{{u}^{2}}-1-2021}{u}.2u\text{d}u}=2\int{\left( {{u}^{2}}-2022 \right)}\text{d}u$.
& \text{d}x=2u\text{d}u \\
& x={{u}^{2}}-1 \\
\end{aligned} \right.$.
.Khi đó $\int{\dfrac{x-2021}{\sqrt{x+1}}\text{d}x}=\int{\dfrac{{{u}^{2}}-1-2021}{u}.2u\text{d}u}=2\int{\left( {{u}^{2}}-2022 \right)}\text{d}u$.
Đáp án C.