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Câu hỏi: Cho hàm số f(x) liên tục trên $\mathbb{R}\backslash \left\{ 0 \right\}$ thỏa mãn $f\left( x \right)+4f\left( \dfrac{1}{x} \right)=8{{x}^{2}}.$ Tính tích phân $I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}.$
A. $I=\dfrac{3}{2}.$
B. $I=\dfrac{9}{2}.$
C. $I=3.$
D. $I=4.$
A. $I=\dfrac{3}{2}.$
B. $I=\dfrac{9}{2}.$
C. $I=3.$
D. $I=4.$
Đặt $x=\dfrac{1}{t}$.
$\begin{aligned}
& \Rightarrow I=\int\limits_{2}^{\dfrac{1}{2}}{t.f\left( \dfrac{1}{t} \right)d\left( \dfrac{1}{t} \right)}=\int\limits_{2}^{\dfrac{1}{2}}{t.f\left( \dfrac{1}{t} \right).\dfrac{-1}{{{t}^{2}}}dt}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{t}.f\left( \dfrac{1}{t} \right)dt}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}.f\left( x \right)dx} \\
& \Rightarrow 5I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}+4\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}.f\left( \dfrac{1}{x} \right)dx}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}\left[ f\left( x \right)+4f\left( \dfrac{1}{x} \right) \right]dx}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}.8{{x}^{2}}dx}=4{{x}^{2}}\left| \begin{aligned}
& ^{2} \\
& _{\dfrac{1}{2}} \\
\end{aligned} \right.=15\Rightarrow I=3. \\
\end{aligned}$
$\begin{aligned}
& \Rightarrow I=\int\limits_{2}^{\dfrac{1}{2}}{t.f\left( \dfrac{1}{t} \right)d\left( \dfrac{1}{t} \right)}=\int\limits_{2}^{\dfrac{1}{2}}{t.f\left( \dfrac{1}{t} \right).\dfrac{-1}{{{t}^{2}}}dt}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{t}.f\left( \dfrac{1}{t} \right)dt}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}.f\left( x \right)dx} \\
& \Rightarrow 5I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}+4\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}.f\left( \dfrac{1}{x} \right)dx}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}\left[ f\left( x \right)+4f\left( \dfrac{1}{x} \right) \right]dx}=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x}.8{{x}^{2}}dx}=4{{x}^{2}}\left| \begin{aligned}
& ^{2} \\
& _{\dfrac{1}{2}} \\
\end{aligned} \right.=15\Rightarrow I=3. \\
\end{aligned}$
Đáp án C.