Cho

B

- Hướng dẫn giải

Đáp án B

$$\begin{gathered} u=e^x+1\Rightarrow du=e^{x}dx \\ \Rightarrow dx=\frac{du}{u-1}. \end{gathered}$$

$$\begin{gathered} \underset{2}{\overset{e+1}{\int }}\frac{du}{u\left(u-1\right)}=\underset{2}{\overset{e+1}{\int }}\left(\frac{1}{u-1}-\frac{1}{u}\right)du \\ ={\left.\ln \left|\frac{u-1}{u}\right|\right|}_2^{e+1}=\ln \frac{2e}{e+1}=-\ln \frac{e+1}{2e} \\ =-\left(\ln \frac{e+1}{2}-\ln e\right)=1-\ln \frac{e+1}{2}. \end{gathered}$$

$a=\mathrm{1,}b=-1\Rightarrow S=a+b=0.$