Cho hàm số f(x) xác định trên (0; dương vô cùng

* Hướng dẫn giải

Đáp án A

Ta có $f^{\text{'}}\left(x\right)=\frac{1}{x\left(\ln x-1\right)}$

$\Rightarrow f\left(x\right)=\int \frac{1}{x\left(\ln x-1\right)}dx=\ln \left|\ln x-1\right|+C=\left\{\begin{array}{l}\ln \left(1-\ln x\right)+C_{1}khix\in \left(0;e\right)\\ \ln \left(\ln x-1\right)+C_{2}khix\in \left(e;+\infty \right)\end{array}\right.$.

+) $f\left(\frac{1}{e^2}\right)=\ln 6\Rightarrow C_1=\ln 2$.

+) $f\left(e^2\right)=3\Rightarrow C_2=3$.

Do đó $f\left(x\right)=\left\{\begin{array}{l}\ln \left(1-\ln x\right)+\ln 2khix\in \left(0;e\right)\\ \ln \left(\ln x-1\right)+3khix\in \left(e;+\infty \right)\end{array}\right.\Rightarrow \left\{\begin{array}{l}f\left(\frac{1}{e}\right)=\ln 2+\ln 2\\ f\left(e^3\right)=\ln 2+3\end{array}\right.$

$\stackrel{}{\to }f\left(\frac{1}{e}\right)+f\left(e^3\right)=3\left(\ln 2+1\right)$.