The integral $\int \left({\left(\frac{\mathrm{x}}{e}\right)}^{\mathrm{x}}+{\left(\frac{e}{\mathrm{x}}\right)}^{\mathrm{x}}\right){\log }_{e}x dx$ is equal to

$\int \left[{\left(\frac{\mathrm{x}}{e}\right)}^{\mathrm{x}}+{\left(\frac{e}{\mathrm{x}}\right)}^{\mathrm{x}}\right]{\log }_{e}x\mathrm{dx}=$
$$\begin{gathered} {\left(\frac{x}{e}\right)}^x=t \\ {\left(\frac{x}{e}\right)}^x\left(x\frac{1}{x}+\log x-1\right)dx=dt \\ {\left(\frac{x}{e}\right)}^x\left(1+\log x-1\right)dx=dt \\ {\left(\frac{x}{e}\right)}^x\log xdx=dt \end{gathered}$$
$$\begin{gathered} {\left(\frac{e}{x}\right)}^x=s \\ {\left(\frac{e}{x}\right)}^x\left(x\left(-\frac{1}{x}\right)+\log \left(\frac{e}{x}\right)\right)dx=ds \\ {\left(\frac{e}{x}\right)}^x\left(-1+1-\log x\right)dx=ds \\ -{\left(\frac{e}{x}\right)}^x\log xdx=ds \end{gathered}$$

 $$\begin{gathered} \int \left[{\left(\frac{\mathrm{x}}{\mathrm{e}}\right)}^{\mathrm{x}}+{\left(\frac{\mathrm{e}}{\mathrm{x}}\right)}^{\mathrm{x}}\right]{\log }_{\mathrm{e}}^{\mathrm{x}}dx \\ =\int dt-\int ds \\ =t-s+C \\ {\left(\frac{\mathrm{x}}{e}\right)}^{\mathrm{x}}-{\left(\frac{e}{\mathrm{x}}\right)}^{\mathrm{x}}+C \end{gathered}$$
*Note: Given question and options are wrong *
$\int \left({\left(\frac{\mathrm{x}}{2}\right)}^{\mathrm{x}}+{\left(\frac{2}{\mathrm{x}}\right)}^{\mathrm{x}}\right){\log }_2^{ }xdx$
${\left(\frac{\mathrm{x}}{2}\right)}^{\mathrm{x}}-{\left(\frac{2}{\mathrm{x}}\right)}^{\mathrm{x}}+\mathrm{C}$