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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)=x{{\left( {{x}^{2}}+1 \right)}^{2022}}$ thỏa mãn $F\left( 0 \right)=\dfrac{1}{4046}$. Giá trị của $F\left( 1 \right)$ bằng:
A. ${{2}^{2023}}$
B. $\dfrac{{{2}^{2023}}}{2023}$
C. ${{2}^{2022}}$
D. $\dfrac{{{2}^{2022}}}{2023}$
$F\left( x \right)=\int{f\left( x \right)dx}=\int{x{{\left( {{x}^{2}}+1 \right)}^{2022}}dx}$
Đặt $t={{x}^{2}}+1\Rightarrow dt=2xdx\Leftrightarrow \dfrac{dt}{2}=xdx$ $$
Khi đó $F\left( x \right)=\int{{{t}^{2022}}\frac{dt}{2}}=\frac{1}{2}.\frac{{{t}^{2023}}}{2023}+C=\frac{{{\left( {{x}^{2}}+1 \right)}^{2023}}}{4046}+C$.
$F\left( 0 \right)=\frac{1}{4046}\Leftrightarrow \frac{1}{4046}+C=\frac{1}{4046}\Leftrightarrow C=0$.
Vậy $F\left( x \right)=\frac{{{\left( {{x}^{2}}+1 \right)}^{2023}}}{4046}\Rightarrow F\left( 1 \right)=\frac{{{2}^{2023}}}{4046}=\frac{{{2}^{2022}}}{2023}$.
A. ${{2}^{2023}}$
B. $\dfrac{{{2}^{2023}}}{2023}$
C. ${{2}^{2022}}$
D. $\dfrac{{{2}^{2022}}}{2023}$
$F\left( x \right)=\int{f\left( x \right)dx}=\int{x{{\left( {{x}^{2}}+1 \right)}^{2022}}dx}$
Đặt $t={{x}^{2}}+1\Rightarrow dt=2xdx\Leftrightarrow \dfrac{dt}{2}=xdx$ $$
Khi đó $F\left( x \right)=\int{{{t}^{2022}}\frac{dt}{2}}=\frac{1}{2}.\frac{{{t}^{2023}}}{2023}+C=\frac{{{\left( {{x}^{2}}+1 \right)}^{2023}}}{4046}+C$.
$F\left( 0 \right)=\frac{1}{4046}\Leftrightarrow \frac{1}{4046}+C=\frac{1}{4046}\Leftrightarrow C=0$.
Vậy $F\left( x \right)=\frac{{{\left( {{x}^{2}}+1 \right)}^{2023}}}{4046}\Rightarrow F\left( 1 \right)=\frac{{{2}^{2023}}}{4046}=\frac{{{2}^{2022}}}{2023}$.
Đáp án D.