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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left[ \dfrac{1}{2};2 \right]$ thỏa mãn $f\left( x \right)+x.f\left( \dfrac{1}{x} \right)={{x}^{3}}+{{x}^{2}}.$ Giá trị tích phân $I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)dx}{{{x}^{2}}+x}}$ thuộc khoảng nào trong các khoảng sau
A. $\left( 0;1 \right).$
B. $\left( 1;2 \right).$
C. $\left( 3;4 \right).$
D. $\left( 4;5 \right).$
A. $\left( 0;1 \right).$
B. $\left( 1;2 \right).$
C. $\left( 3;4 \right).$
D. $\left( 4;5 \right).$
Ta có: $f\left( x \right)+x.f\left( \dfrac{1}{x} \right)={{x}^{3}}+{{x}^{2}}\Leftrightarrow \dfrac{f\left( x \right)}{{{x}^{2}}+x}+\dfrac{1}{x+1}f\left( \dfrac{1}{x} \right)=x$
Lấy tích phân 2 vế cận từ $\dfrac{1}{2}\to 2$ ta có: $\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}}+\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x+1}f\left( \dfrac{1}{x} \right)=\int\limits_{\dfrac{1}{2}}^{2}{xdx}=\dfrac{15}{8}}$ (*)
Đặt $t=\dfrac{1}{x}\Rightarrow dt=\dfrac{-1}{{{x}^{2}}}dx=-{{t}^{2}}dx\Leftrightarrow dx=\dfrac{-dt}{{{t}^{2}}}$
Thực hiện phép đổi cận ta có: $\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x+1}f\left( \dfrac{1}{x} \right)=\int\limits_{2}^{\dfrac{1}{2}}{\dfrac{1}{\dfrac{1}{t}+1}f\left( t \right).\dfrac{-dt}{{{t}^{2}}}=-\int\limits_{2}^{\dfrac{1}{2}}{\dfrac{1}{{{t}^{2}}+t}f\left( t \right)dt}}}$
$=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( t \right)}{{{t}^{2}}+t}dt=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)dx}{{{x}^{2}}+x}.}}$
Vậy (*) $\Leftrightarrow 2I=\dfrac{15}{8}\Leftrightarrow I=\dfrac{15}{16}.$
Lấy tích phân 2 vế cận từ $\dfrac{1}{2}\to 2$ ta có: $\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}}+\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x+1}f\left( \dfrac{1}{x} \right)=\int\limits_{\dfrac{1}{2}}^{2}{xdx}=\dfrac{15}{8}}$ (*)
Đặt $t=\dfrac{1}{x}\Rightarrow dt=\dfrac{-1}{{{x}^{2}}}dx=-{{t}^{2}}dx\Leftrightarrow dx=\dfrac{-dt}{{{t}^{2}}}$
Thực hiện phép đổi cận ta có: $\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{1}{x+1}f\left( \dfrac{1}{x} \right)=\int\limits_{2}^{\dfrac{1}{2}}{\dfrac{1}{\dfrac{1}{t}+1}f\left( t \right).\dfrac{-dt}{{{t}^{2}}}=-\int\limits_{2}^{\dfrac{1}{2}}{\dfrac{1}{{{t}^{2}}+t}f\left( t \right)dt}}}$
$=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( t \right)}{{{t}^{2}}+t}dt=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)dx}{{{x}^{2}}+x}.}}$
Vậy (*) $\Leftrightarrow 2I=\dfrac{15}{8}\Leftrightarrow I=\dfrac{15}{16}.$
Đáp án A.