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Câu hỏi: Biết $F(x)$ là một nguyên hàm của $f(x)=\dfrac{{{\left[ \ln \left( x+\sqrt{{{x}^{2}}+1} \right) \right]}^{2022}}}{\sqrt{{{x}^{2}}+1}}$ và $F(0)=1$. Giá trị của $F(1)$ bằng
A. $\dfrac{{{\left[ \ln \left( 1+\sqrt{2} \right) \right]}^{2023}}-2023}{2023}$.
B. $\dfrac{{{\left[ \ln \left( 1+\sqrt{2} \right) \right]}^{2023}}+2023}{2023}$.
C. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2023}}-2023}{2023}$.
D. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2023}}+2023}{2023}$.
A. $\dfrac{{{\left[ \ln \left( 1+\sqrt{2} \right) \right]}^{2023}}-2023}{2023}$.
B. $\dfrac{{{\left[ \ln \left( 1+\sqrt{2} \right) \right]}^{2023}}+2023}{2023}$.
C. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2023}}-2023}{2023}$.
D. $\dfrac{{{\left( 1+\sqrt{2} \right)}^{2023}}+2023}{2023}$.
Ta có $F(x)=\int{\dfrac{{{\left[ \ln \left( x+\sqrt{{{x}^{2}}+1} \right) \right]}^{2022}}}{\sqrt{{{x}^{2}}+1}}}dx$. Đặt $t=\ln \left( x+\sqrt{{{x}^{2}}+1} \right)\Rightarrow dt=\dfrac{1}{\sqrt{{{x}^{2}}+1}}dx$
Vậy $F(x)=\int{{{t}^{2022}}dt}=\dfrac{{{t}^{2023}}}{2023}+C=\dfrac{{{\left[ \ln \left( x+\sqrt{{{x}^{2}}+1} \right) \right]}^{2023}}}{2023}+C$
Theo giả thiết ta có
$F(0)=1\Leftrightarrow C=1\Rightarrow F(x)=\dfrac{{{\left[ \ln \left( x+\sqrt{{{x}^{2}}+1} \right) \right]}^{2023}}}{2023}+1\Rightarrow F(1)=\dfrac{{{\left[ \ln \left( 1+\sqrt{2} \right) \right]}^{2023}}+2023}{2023}$.
Vậy $F(x)=\int{{{t}^{2022}}dt}=\dfrac{{{t}^{2023}}}{2023}+C=\dfrac{{{\left[ \ln \left( x+\sqrt{{{x}^{2}}+1} \right) \right]}^{2023}}}{2023}+C$
Theo giả thiết ta có
$F(0)=1\Leftrightarrow C=1\Rightarrow F(x)=\dfrac{{{\left[ \ln \left( x+\sqrt{{{x}^{2}}+1} \right) \right]}^{2023}}}{2023}+1\Rightarrow F(1)=\dfrac{{{\left[ \ln \left( 1+\sqrt{2} \right) \right]}^{2023}}+2023}{2023}$.
Đáp án B.
