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