$\int {{e^x} \left[ \frac{1 + x \log x}{x} \right] \, dx} = $

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

$\int e^{-x}(x^3-2x^2+3x-4) dx=$

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

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

If $\int {{e^{\sec x}}\left( {\sec x + \tan x f(x) + (\sec x \tan x + \sec^2 x)} \right)dx = {e^{\sec x}}f(x) + C}$,then a possible choice of $f(x)$ is