$\int e^x(1-\cot x+\cot^2 x) dx =$

If $\int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx = a+\frac{b}{\log 2}$,then:

$\int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx = $

$\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$,where $C$ is a constant,then $\frac{d^{3} f}{d x^{3}}$ at $x = 1$ is equal to

$\int e^x \left( \log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^2} \right) dx$ is not the original,the question is $\int e^x \left( \log x + \frac{1}{x} \right) dx$ or similar. Given the options,the integral is $\int e^x \left( \log x + \frac{1}{x} \right) dx$. Let us evaluate $\int e^x \left( \log x + \frac{1}{x} \right) dx$.

$\int \frac{\log _e x}{\left(1+\log _e x\right)^2} d x=$