Biết

B

- Hướng dẫn giải

Đáp án B

$$\begin{gathered} I={\int }_1^5\frac{2\left|x-2\right|+1}{x}dx={\int }_1^2\frac{2\left|x-2\right|+1}{x}dx+{\int }_2^5\frac{2\left|x-2\right|+1}{x}dx={\int }_1^2\frac{2\left(x-2\right)+1}{x}dx+{\int }_2^5\frac{2\left(x-2\right)+1}{x}dx \\ ={\int }_1^2\left(\frac{5}{x}-2\right)dx+{\int }_2^5\left(2-\frac{3}{x}\right)dx={\overline{)\left(5\ln \left|x\right|-2x\right)}}_1^2+{\overline{)\left(2x-3\ln \left|x\right|\right)}}_2^5=8\ln 2-2\ln 5+4\Rightarrow \left\{\begin{array}{l}a=8\\ b=-3\end{array}\Rightarrow a-b=11\right. \end{gathered}$$