Google
Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)=\dfrac{2x+1}{{{x}^{4}}+2{{x}^{3}}+{{x}^{2}}}$ trên khoảng $\left( 0;+\infty \right)$ thỏa mãn $F\left( \dfrac{1}{2} \right)=\dfrac{6059}{3}$. Tính $F\left( 1 \right)$.
A. 2021.
B. $\dfrac{4021}{2}$.
C. $\dfrac{4041}{2020}$.
D. 2020.
A. 2021.
B. $\dfrac{4021}{2}$.
C. $\dfrac{4041}{2020}$.
D. 2020.
Ta có $f\left( x \right)=\dfrac{2x+1}{{{x}^{4}}+2{{x}^{3}}+{{x}^{2}}}=\dfrac{2x+1}{{{\left( {{x}^{2}}+x \right)}^{2}}}$.
Đặt $t={{x}^{2}}+x\Rightarrow dt=\left( 2x+1 \right)dx$.
Khi đó $F\left( x \right)=\int{f\left( x \right)dx}=\int{\dfrac{1}{{{t}^{2}}}dt}=-\dfrac{1}{t}+C=-\dfrac{1}{\left( {{x}^{2}}+x \right)}+C$.
Mặt khác, $F\left( \dfrac{1}{2} \right)=\dfrac{6059}{3}\Rightarrow -\dfrac{4}{3}+C=\dfrac{6059}{3}\Rightarrow C=2021$.
Vậy $F\left( x \right)=-\dfrac{1}{\left( {{x}^{2}}+x \right)}+2021\Rightarrow F\left( 1 \right)=\dfrac{4041}{2}$.
Đặt $t={{x}^{2}}+x\Rightarrow dt=\left( 2x+1 \right)dx$.
Khi đó $F\left( x \right)=\int{f\left( x \right)dx}=\int{\dfrac{1}{{{t}^{2}}}dt}=-\dfrac{1}{t}+C=-\dfrac{1}{\left( {{x}^{2}}+x \right)}+C$.
Mặt khác, $F\left( \dfrac{1}{2} \right)=\dfrac{6059}{3}\Rightarrow -\dfrac{4}{3}+C=\dfrac{6059}{3}\Rightarrow C=2021$.
Vậy $F\left( x \right)=-\dfrac{1}{\left( {{x}^{2}}+x \right)}+2021\Rightarrow F\left( 1 \right)=\dfrac{4041}{2}$.
Đáp án B.