The integrating factor of the differential equation $\left(x\log \mathrm{_e}x\right)\frac{d}{y}/\frac{d}{x}+y=2\log \mathrm{_e}x$ is

The correct option is:
$\log e_{}x$

To find the integrating factor of the given differential equation, we first write it in standard linear form. The given differential equation is:
$\left(x\log e_{}x\right)\frac{dy}{dx}+y=2\log e_{}x$

A first-order linear differential equation is typically written in the standard form:
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

To convert our equation into this standard form, we divide every term by the coefficient of $\frac{dy}{dx}$, which is $x\log e_{}x$:
$\frac{dy}{dx}+\frac{1}{x\log e_{}x}y=\frac{2\log e_{}x}{x\log e_{}x}$

Comparing this with the standard form, we identify the coefficient function $P\left(x\right)$ as:
$P\left(x\right)=\frac{1}{x\log e_{}x}$

The formula for the integrating factor (I.F.) is:
$\text{I.F.}=e^{\int P\left(x\right)dx}$

First, we evaluate the integral:
$\int P\left(x\right)dx=\int \frac{1}{x\log e_{}x}dx$

To solve this integral, we can use the substitution method. Let:
$u=\log e_{}x$
Differentiating both sides with respect to $x$ gives:
$du=\frac{1}{x}dx$

Substituting these values back into the integral:

Replacing $u$ with its original value $\log e_{}x$ yields:
$\int P\left(x\right)dx=\log e_{}\left(\log e_{}x\right)$

Now we calculate the integrating factor:
$\text{I.F.}=e^{\log e_{}\left(\log e_{}x\right)}$

Using the exponential-logarithmic identity $e^{\log e_{}f\left(x\right)}=f\left(x\right)$, the expression simplifies directly to:
$\text{I.F.}=\log e_{}x$