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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 tích phân $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx}$ bằng
A. $\dfrac{8}{9}$.
B. $\dfrac{2}{3}$.
C. $\dfrac{3}{4}$.
D. $\dfrac{16}{9}$.
A. $\dfrac{8}{9}$.
B. $\dfrac{2}{3}$.
C. $\dfrac{3}{4}$.
D. $\dfrac{16}{9}$.
+ Đặt $x=\dfrac{1}{t}\Rightarrow dx=-\dfrac{1}{{{t}^{2}}}dt.$
+ Đổi cận: $x=\dfrac{1}{3}\Rightarrow t=3;x=3\Rightarrow t=\dfrac{1}{3}.$
+ Ta có $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx=-\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{f\left( \dfrac{1}{t} \right)}{\dfrac{1}{{{t}^{2}}}+\dfrac{1}{t}}.\dfrac{1}{{{t}^{2}}}dt=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{t} \right)}{t+1}dt.}}}$
Suy ra: $2I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)+x.f\left( \dfrac{1}{x} \right)}{x\left( x+1 \right)}dx}}$
$=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{x.\left( x-1 \right)\left( x+1 \right)}{x\left( x+1 \right)}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)dx}=\dfrac{16}{9}.$
Vậy $I=\dfrac{8}{9}.$
Đặt $u=3-{{x}^{2}}\Rightarrow du=-2xdx$ với $\left\{ \begin{aligned}
& x=-1\Rightarrow u=2 \\
& x=2\Rightarrow u=-1 \\
\end{aligned} \right.$
Khi đó $\int\limits_{-1}^{2}{xf\left( 3-{{x}^{2}} \right)dx}=\dfrac{1}{2}\int\limits_{-1}^{2}{f\left( u \right)du}=\dfrac{1}{2}\int\limits_{-1}^{2}{f\left( x \right)}dx$ thay vào (*) ta được
$\int\limits_{-1}^{2}{f\left( x \right)dx}-\dfrac{1}{2}\int\limits_{-1}^{2}{f\left( x \right)dx}=\dfrac{14}{3}\Leftrightarrow \int\limits_{-1}^{2}{f\left( x \right)dx}=\dfrac{28}{3}.$
+ Đổi cận: $x=\dfrac{1}{3}\Rightarrow t=3;x=3\Rightarrow t=\dfrac{1}{3}.$
+ Ta có $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx=-\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{f\left( \dfrac{1}{t} \right)}{\dfrac{1}{{{t}^{2}}}+\dfrac{1}{t}}.\dfrac{1}{{{t}^{2}}}dt=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{t} \right)}{t+1}dt.}}}$
Suy ra: $2I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)+x.f\left( \dfrac{1}{x} \right)}{x\left( x+1 \right)}dx}}$
$=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{x.\left( x-1 \right)\left( x+1 \right)}{x\left( x+1 \right)}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)dx}=\dfrac{16}{9}.$
Vậy $I=\dfrac{8}{9}.$
Đặt $u=3-{{x}^{2}}\Rightarrow du=-2xdx$ với $\left\{ \begin{aligned}
& x=-1\Rightarrow u=2 \\
& x=2\Rightarrow u=-1 \\
\end{aligned} \right.$
Khi đó $\int\limits_{-1}^{2}{xf\left( 3-{{x}^{2}} \right)dx}=\dfrac{1}{2}\int\limits_{-1}^{2}{f\left( u \right)du}=\dfrac{1}{2}\int\limits_{-1}^{2}{f\left( x \right)}dx$ thay vào (*) ta được
$\int\limits_{-1}^{2}{f\left( x \right)dx}-\dfrac{1}{2}\int\limits_{-1}^{2}{f\left( x \right)dx}=\dfrac{14}{3}\Leftrightarrow \int\limits_{-1}^{2}{f\left( x \right)dx}=\dfrac{28}{3}.$
Đáp án A.