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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left[ \dfrac{1}{3};3 \right]$ thỏa mãn $f\left( x \right)+x.f\left( \dfrac{1}{x} \right)={{x}^{3}}-x.$ Giá trị của tích phân $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}}dx$ bằng
A. $\dfrac{16}{9}.$
B. $\dfrac{2}{3}.$
C. $\dfrac{3}{4}.$
D. $\dfrac{8}{9}.$
A. $\dfrac{16}{9}.$
B. $\dfrac{2}{3}.$
C. $\dfrac{3}{4}.$
D. $\dfrac{8}{9}.$
Đặt $t=\dfrac{1}{x}\Rightarrow I=\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{f\left( \dfrac{1}{t} \right)}{{{\left( \dfrac{1}{t} \right)}^{2}}+\dfrac{1}{t}}d\left( \dfrac{1}{t} \right)=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{t} \right)}{\dfrac{1}{{{t}^{2}}}+\dfrac{1}{t}}.\dfrac{1}{{{t}^{2}}}dt=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{t} \right)}{t+1}dt}}}$
$\Rightarrow I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{x.f\left( \dfrac{1}{x} \right)}{{{x}^{2}}+x}dx}\Rightarrow I+I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)+x.f\left( \dfrac{1}{x} \right)}{{{x}^{2}}+x}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{{{x}^{3}}-x}{{{x}^{2}}+x}dx}$
$=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{x\left( x+1 \right)\left( x-1 \right)}{x\left( x+1 \right)}dx}=\left( \dfrac{{{x}^{2}}}{2}-x \right)\left| _{\begin{smallmatrix}
\\
\dfrac{1}{3}
\end{smallmatrix}}^{\begin{smallmatrix}
3 \\
\end{smallmatrix}} \right.=\dfrac{16}{9}\Rightarrow I=\dfrac{8}{9}.$
$\Rightarrow I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{x.f\left( \dfrac{1}{x} \right)}{{{x}^{2}}+x}dx}\Rightarrow I+I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)+x.f\left( \dfrac{1}{x} \right)}{{{x}^{2}}+x}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{{{x}^{3}}-x}{{{x}^{2}}+x}dx}$
$=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{x\left( x+1 \right)\left( x-1 \right)}{x\left( x+1 \right)}dx}=\left( \dfrac{{{x}^{2}}}{2}-x \right)\left| _{\begin{smallmatrix}
\\
\dfrac{1}{3}
\end{smallmatrix}}^{\begin{smallmatrix}
3 \\
\end{smallmatrix}} \right.=\dfrac{16}{9}\Rightarrow I=\dfrac{8}{9}.$
Đáp án D.