- Hướng dẫn giải

$$\begin{gathered} {\int }_0^{\ln 2}\left(x+\frac{1}{2e^x+1}\right)dx={\int }_0^{\ln 2}xdx+{\int }_0^{\ln 2}\frac{1}{2e^x+1}dx \\ ={\frac{x^2}{2}}_0^{\ln 2}+{\int }_0^{\ln 2}\frac{1}{\left(2e^x+1\right)e^x}d{e}^x \end{gathered}$$

= $$\begin{gathered} \frac{1}{2}{\ln }^{2}2+{\int }_0^{\ln 2}\left(\frac{1}{e^x}-\frac{2}{\left(2e^x+1\right)}\right)d{e}^x \\ =\frac{1}{2}{\ln }^{2}2+{\left(\ln e^x-\ln \left(2e^x+1\right)\right)}_0^{\ln 2} \\ =\frac{1}{2}{\ln }^{2}2+\ln 2-\ln \frac{5}{3} \end{gathered}$$

Vậy a + b + c = 2

Đáp án cần chọn là A