Cho hàm số $\Large f(x)=\left\{\begin{align}&\dfrac{1}{2}x+2\ khi 0\le

Cho hàm số $\Large f(x)=\left\{\begin{align}&\dfrac{1}{2}x+2\ khi 0\leq x < 2\\&-x+5\ khi 2\leq x\leq 5\\\end{align}\right.$. Khi đó $\Large \int_1^{e^{2}}\dfrac{f(\ln x}{x}dx+\int_{\sqrt{3}}{2\sqrt{6}}xf(\sqrt{x^{2}+1})dx$ bằng

• A. $\Large \dfrac{37}{2}$
• B. $\Large 5$
• C. $\Large \dfrac{19}{2}$
• D. $\Large \dfrac{27}{2}$

Ta có

$\Large +)I_1=\int_1^{e^{2}}\dfrac{f(\ln x)}{x}dx=\int_1^{e^{2}}f(\ln x)\Leftrightarrow I_1=\int_0^{2}f(u)du$

$\Large \Leftrightarrow I_1=\int_0^{2}f(x)dx=\int_0^{2}\left(\dfrac{1}{2}x+2\right)dx=5$

$\Large +)I_2=\int_{\sqrt{3}}{2\sqrt{6}}xf(\sqrt{x^{2}+1})dx=\int_\sqrt{3}^{2\sqrt{6}}\sqrt{x^{2}+1}.f(\sqrt{x^{2}+1})d(\sqrt{x^{2}+1}$

$\Large \Leftrightarrow I_2=\int_2^{5}u.f(u)du=\int_2^{5}x.f(x)dx\Leftrightarrow I_2=\int_2^{5}x.(-x+5)dx=\dfrac{27}{2}$

Vậy $\Large \int_1^{e^{2}}\dfrac{f(\ln x)}{x}dx+\int_\sqrt{3}^{2\sqrt{6}}xf(\sqrt{x^{2}+1})dx+\int_\sqrt{3}^{2\sqrt{6}}xf(\sqrt{x^{2}+1})dx=5+\dfrac{27}{2}=\dfrac{37}{2}$