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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)=\dfrac{{{\left( x+\sqrt{{{x}^{2}}+1} \right)}^{2021}}}{\sqrt{{{x}^{2}}+1}}$ và $F\left( 0 \right)=1.$ Giá trị của $F\left( 1 \right)$ bằng
A. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2020}}-2021}{2020}$
B. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2021}}+2020}{2021}$
C. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2020}}+2021}{2020}$
D. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2021}}-2020}{2021}$
A. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2020}}-2021}{2020}$
B. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2021}}+2020}{2021}$
C. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2020}}+2021}{2020}$
D. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2021}}-2020}{2021}$
Cách giải:
Đặt $x+\sqrt{{{x}^{2}}+1}=t\Rightarrow \left( 1+\dfrac{x}{\sqrt{{{x}^{2}}+1}} \right)dx=dt$
$\Rightarrow \dfrac{\left( \sqrt{{{x}^{2}}+1}+x \right)}{\sqrt{{{x}^{2}}+1}}dx=dt\Rightarrow \dfrac{t}{\sqrt{{{x}^{2}}+1}}dx=dt\Rightarrow \dfrac{dx}{\sqrt{{{x}^{2}}+1}}=\dfrac{dt}{t}$
Khi đó: $x:0\to 1\Rightarrow t:1\to 1+\sqrt{2}$
$\int\limits_{0}^{1}{\dfrac{{{\left( x+\sqrt{{{x}^{2}}+1} \right)}^{2021}}}{\sqrt{{{x}^{2}}+1}}dx}=\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{t}^{2021}}}{t}dt}=\int\limits_{1}^{1+\sqrt{2}}{{{t}^{2020}}dt}=\dfrac{{{t}^{2020}}}{2021}\left| \begin{aligned}
& 1+\sqrt{2} \\
& 1 \\
\end{aligned} \right.=\dfrac{1}{2021}.\left[ {{\left( 1+\sqrt{2} \right)}^{2021}}-1 \right]$
$=F\left( 1 \right)-F\left( 0 \right)\Rightarrow F\left( 1 \right)=\dfrac{1}{2021}.\left[ {{\left( 1+\sqrt{2} \right)}^{2021}}-1 \right]+1=\dfrac{{{\left( 1+\sqrt{2} \right)}^{2021}}+2020}{2021}$
Đặt $x+\sqrt{{{x}^{2}}+1}=t\Rightarrow \left( 1+\dfrac{x}{\sqrt{{{x}^{2}}+1}} \right)dx=dt$
$\Rightarrow \dfrac{\left( \sqrt{{{x}^{2}}+1}+x \right)}{\sqrt{{{x}^{2}}+1}}dx=dt\Rightarrow \dfrac{t}{\sqrt{{{x}^{2}}+1}}dx=dt\Rightarrow \dfrac{dx}{\sqrt{{{x}^{2}}+1}}=\dfrac{dt}{t}$
Khi đó: $x:0\to 1\Rightarrow t:1\to 1+\sqrt{2}$
$\int\limits_{0}^{1}{\dfrac{{{\left( x+\sqrt{{{x}^{2}}+1} \right)}^{2021}}}{\sqrt{{{x}^{2}}+1}}dx}=\int\limits_{1}^{1+\sqrt{2}}{\dfrac{{{t}^{2021}}}{t}dt}=\int\limits_{1}^{1+\sqrt{2}}{{{t}^{2020}}dt}=\dfrac{{{t}^{2020}}}{2021}\left| \begin{aligned}
& 1+\sqrt{2} \\
& 1 \\
\end{aligned} \right.=\dfrac{1}{2021}.\left[ {{\left( 1+\sqrt{2} \right)}^{2021}}-1 \right]$
$=F\left( 1 \right)-F\left( 0 \right)\Rightarrow F\left( 1 \right)=\dfrac{1}{2021}.\left[ {{\left( 1+\sqrt{2} \right)}^{2021}}-1 \right]+1=\dfrac{{{\left( 1+\sqrt{2} \right)}^{2021}}+2020}{2021}$
Đáp án B.