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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 $f\left( x \right)+x\cdot f\left( \dfrac{1}{x} \right)={{x}^{3}}-x,\forall x\in \left[ \dfrac{1}{3};3 \right]$. Tích phân $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f(x)}{{{x}^{2}}+x}\text{d}x}$ bằng
A. $\dfrac{2}{3}$.
B. $\dfrac{16}{9}$.
C. $\dfrac{8}{9}$.
D. $\dfrac{3}{4}$.
A. $\dfrac{2}{3}$.
B. $\dfrac{16}{9}$.
C. $\dfrac{8}{9}$.
D. $\dfrac{3}{4}$.
Ta có $f\left( x \right)+x\cdot f\left( \dfrac{1}{x} \right)={{x}^{3}}-x\Leftrightarrow \dfrac{f\left( x \right)}{{{x}^{2}}+x}+\dfrac{1}{x+1}\cdot f\left( \dfrac{1}{x} \right)=x-1$.
Suy ra $\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f(x)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{1}{x+1}\cdot f\left( \dfrac{1}{x} \right)\text{d}x}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)\text{d}x}.\left( * \right)$
Đặt $t=\dfrac{1}{x}$ suy ra $\text{d}x=-\dfrac{1}{{{t}^{2}}}\text{d}t$ và $x=\dfrac{1}{t}\Leftrightarrow \dfrac{1}{x+1}=\dfrac{t}{t+1}$.
Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{1}{3}\Rightarrow t=3 \\
& x=3\Rightarrow t=\dfrac{1}{3} \\
\end{aligned} \right.$. Khi đó
$\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{1}{x+1}\cdot f\left( \dfrac{1}{x} \right)\text{d}x}=-\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{t}{t+1}\cdot f\left( t \right)\cdot \dfrac{1}{{{t}^{2}}}\text{d}t}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( t \right)}{{{t}^{2}}+t}\text{d}t}=I$.
Do đó $\left( * \right)\Leftrightarrow 2I=\dfrac{16}{9}\Leftrightarrow I=\dfrac{8}{9}$.
Suy ra $\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f(x)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{1}{x+1}\cdot f\left( \dfrac{1}{x} \right)\text{d}x}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)\text{d}x}.\left( * \right)$
Đặt $t=\dfrac{1}{x}$ suy ra $\text{d}x=-\dfrac{1}{{{t}^{2}}}\text{d}t$ và $x=\dfrac{1}{t}\Leftrightarrow \dfrac{1}{x+1}=\dfrac{t}{t+1}$.
Đổi cận $\left\{ \begin{aligned}
& x=\dfrac{1}{3}\Rightarrow t=3 \\
& x=3\Rightarrow t=\dfrac{1}{3} \\
\end{aligned} \right.$. Khi đó
$\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{1}{x+1}\cdot f\left( \dfrac{1}{x} \right)\text{d}x}=-\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{t}{t+1}\cdot f\left( t \right)\cdot \dfrac{1}{{{t}^{2}}}\text{d}t}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( t \right)}{{{t}^{2}}+t}\text{d}t}=I$.
Do đó $\left( * \right)\Leftrightarrow 2I=\dfrac{16}{9}\Leftrightarrow I=\dfrac{8}{9}$.
Đáp án C.