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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& \dfrac{1}{2}x+2\text{ khi }0\le x\le 2 \\
& -x+5\text{ khi }2\le x\le 5 \\
\end{aligned} \right. $. Khi đó $ \int\limits_{1}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}dx}+\int\limits_{\sqrt{3}}^{2\sqrt{6}}{x.f\left( \sqrt{{{x}^{2}}+1} \right)dx}$ bằng
A. $\dfrac{19}{2}$
B. $\dfrac{37}{2}$
C. $\dfrac{27}{2}$
D. 5
& \dfrac{1}{2}x+2\text{ khi }0\le x\le 2 \\
& -x+5\text{ khi }2\le x\le 5 \\
\end{aligned} \right. $. Khi đó $ \int\limits_{1}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}dx}+\int\limits_{\sqrt{3}}^{2\sqrt{6}}{x.f\left( \sqrt{{{x}^{2}}+1} \right)dx}$ bằng
A. $\dfrac{19}{2}$
B. $\dfrac{37}{2}$
C. $\dfrac{27}{2}$
D. 5
Ta có $\int\limits_{1}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}dx}=\int\limits_{1}^{{{e}^{2}}}{f\left( \ln x \right)d\left( \ln x \right)}=\int\limits_{0}^{2}{f\left( t \right)dt}=\int\limits_{0}^{2}{f\left( x \right)dx}=\int\limits_{0}^{2}{\left( \dfrac{1}{2}x+2 \right)dx}$
Lại có $\int\limits_{\sqrt{3}}^{2\sqrt{6}}{x.f\left( \sqrt{{{x}^{2}}+1} \right)dx}=\dfrac{1}{2}\int\limits_{\sqrt{3}}^{2\sqrt{6}}{\sqrt{{{x}^{2}}+1}.f\left( \sqrt{{{x}^{2}}+1} \right)d\left( \sqrt{{{x}^{2}}+1} \right)}=\dfrac{1}{2}\int\limits_{2}^{5}{t.f\left( t \right)dt}=\dfrac{1}{2}\int\limits_{2}^{5}{x.\left( -x+5 \right)dx}$
Vậy $\int\limits_{1}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}dx}+\int\limits_{\sqrt{3}}^{2\sqrt{6}}{x.f\left( \sqrt{{{x}^{2}}+1} \right)dx}=\dfrac{37}{2}$.
Lại có $\int\limits_{\sqrt{3}}^{2\sqrt{6}}{x.f\left( \sqrt{{{x}^{2}}+1} \right)dx}=\dfrac{1}{2}\int\limits_{\sqrt{3}}^{2\sqrt{6}}{\sqrt{{{x}^{2}}+1}.f\left( \sqrt{{{x}^{2}}+1} \right)d\left( \sqrt{{{x}^{2}}+1} \right)}=\dfrac{1}{2}\int\limits_{2}^{5}{t.f\left( t \right)dt}=\dfrac{1}{2}\int\limits_{2}^{5}{x.\left( -x+5 \right)dx}$
Vậy $\int\limits_{1}^{{{e}^{2}}}{\dfrac{f\left( \ln x \right)}{x}dx}+\int\limits_{\sqrt{3}}^{2\sqrt{6}}{x.f\left( \sqrt{{{x}^{2}}+1} \right)dx}=\dfrac{37}{2}$.
Đáp án D.