In the past time I was trying to find a common answer for $\displaystyle\int x^{\ln(x)-1}\ln(x)\,\mathrm dx$, I actually answered the question which pattern is as follows.
$\begin{align*} \displaystyle\int x^{\ln(x)-1}\ln(x)\,\mathrm dx&=\int x^{\ln(x)}x^{-1}\ln(x)\,\mathrm dx \tag*{$\alpha^{m+n}=\alpha^m\alpha^n$} \\ &=\int \frac{x^{\ln(x)}\ln(x)}{x}\,\mathrm dx\\ &=\int\frac{\left( \left( e^{\ln(x)} \right)^{\ln(x)} \right)\ln(x)}{x}\,\mathrm dx\\ &=\int\frac{e^{\ln^2(x)}\ln(x)}{x}\,\mathrm dx \end{align*}\tag*{1}$ Use the Integration by substitution Let $\begin{aligned} u&=\ln(x) \\\dfrac{\mathrm{d}u}{\mathrm{d}x}&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \ln(x) \right] \\ \mathrm du &=\frac{1}{x}\,\mathrm dx \iff\mathrm dx=x\,\mathrm du \end{aligned}\tag*{2}$ $\text{We obtain $\displaystyle\int\frac{e^{\ln^2(x)}\ln(x)}{x}\,\mathrm dx =\int ue^{u^2}\,\mathrm du$}\tag*{3}$ Apply the integration by substitution twice and this time I would use $\alpha$ since I have used $u$ previously. $\begin{aligned} \alpha&=u^2\\\dfrac{\mathrm d\alpha}{\mathrm du} &=\dfrac{\mathrm d}{\mathrm du}\left[ u^2 \right] \\ \mathrm d\alpha&= 2u\,\mathrm du\implies\mathrm du=\frac{1}{2u}\,\mathrm d\alpha \end{aligned}\tag*{4}$ Hence $\begin{aligned} \displaystyle\int\frac{e^{\ln^2(x)}\ln(x)}{x}\,\mathrm dx =\int ue^{u^2}\,\mathrm du\implies\int ue^{u^2}\,\mathrm du&= \frac12\int e^{\alpha}\,\mathrm d\alpha \\ &=\frac{1}{2}e^{\alpha}+C\\ &=\frac{e^{u^2}}{2}+C\qquad \alpha\implies u\\ &=\frac{e^{{\ln^2(x)}}}{2}+C \qquad u\implies x \\&=\frac{e^{\ln(x)\ln(x)}}{2}+C\\ &=\frac{\left( e^{\ln(x)} \right)^{\ln(x)}}{2}+C\\&= \boxed{\frac{x^{\ln(x)}}{2}+C} \end{aligned}\tag*{5}$ Use logarithmic differentiation to tell that $\frac{e^{{\ln^2(x)}}}{2}+C$ is correct. Let $y=\frac{e^{{\ln^2(x)}}}{2}$. $\begin{aligned} y&=e^{\ln^2(x)}\\ \ln(y) &=\ln\left( e^{\ln^2(x)} \right) \\ \ln(y) &=\ln^2(x)\ln(e)\\ \ln(y) &=\ln^2(x)\\\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \ln(y) \right] &=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \ln^2(x) \right]\\ \dfrac{\mathrm{d}}{\mathrm{d}y}\left[ \ln(y) \right]\dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{\mathrm{d}}{\mathrm{d}\alpha}\left[\alpha^2\right]\dfrac{\mathrm{d}}{\mathrm{d}x}[\ln(x)]\\ \frac1y\dfrac{\mathrm{d}y}{\mathrm{d}x}&=2\alpha\left( \frac{1}{x} \right)\\\frac1y\dfrac{\mathrm{d}y}{\mathrm{d}x}&=\frac{2\ln(x)}{x}\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=y\left( \frac{2\ln(x)}{x} \right)\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\frac{2e^{\ln^2(x)}\ln(x)}{x} \end{aligned}\tag*{6}$ Since we have the $\frac{e^{\ln^2(x)}}{2}$, then we can say that like this $\begin{aligned}\displaystyle y=\frac{e^{\ln^2(x)}}{2}&\implies\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \frac{e^{\ln^2(x)}}{2} \right]\\&\implies\frac12\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ e^{\ln^2(x)} \right]\\&\implies\frac{1}{2}\left( \frac{2e^{\ln^2(x)}\ln(x)}{x} \right)\\& \implies\frac{e^{\ln^2(x)}\ln(x)}{x} \\ &\implies\frac{e^{\ln(x)\ln(x)}\ln(x)}{x}\\ &\implies\frac{\left( e^{\ln(x)} \right)^{\ln(x))}\ln(x)}{x}\\&\implies\frac{x^{\ln(x)}\ln(x)}{x}\\ &\implies x^{\ln(x)-1}\ln(x)\end{aligned}\tag*{7}$ The second version is by substituting $u=x^{\ln(x)}$. $\displaystyle\int x^{\ln(x)-1}\ln(x)\,\mathrm dx \tag*{8}$ Let $\begin{aligned} u&=x^{\ln(x)}\\ \frac{\mathrm du}{\mathrm dx}&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ x^{\ln(x)} \right]^* \\\mathrm du &= 2x^{\ln(x)-1}\ln(x)\,\mathrm dx \implies\frac{\mathrm{d}u}{2x^{\ln(x)-1}\ln(x)}=\,\mathrm dx \end{aligned}\tag*{9}$ Then $\begin{aligned} \displaystyle \int x^{\ln(x)-1}\ln(x)\,\mathrm dx&=\int x^{\ln(x)-1}\ln(x)\cdot\frac{1}{2x^{\ln(x)-1}\ln(x)}\,\mathrm du\\ &=\frac12\int 1\,\mathrm du\\ &=\frac12[u]\\ &=\frac{x^{\ln(x)}}{2}+C \end{aligned}\tag*{10}$ *Proof $\frac{\mathrm{d}}{\mathrm{d}x}\left[ x^{\ln(x)} \right]=2x^{\ln(x)-1}\ln(x)$ To proof this let $y=x^{\ln(x)}$ $\begin{aligned} y&=x^{\ln(x)} \\ \ln(y)&=\ln\left( x^{\ln(x)} \right)\\ \ln(y)&=\ln(x)\ln(x)\\ \ln(y)&=\ln^2(x)\\\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \ln(y) \right] &=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \ln^2(x) \right]\\ \dfrac{\mathrm{d}}{\mathrm{d}y}\left[ \ln(y) \right]\dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{\mathrm{d}}{\mathrm{d}\alpha}\left[\alpha^2\right]\dfrac{\mathrm{d}}{\mathrm{d}x}[\ln(x)]\\ \frac1y\dfrac{\mathrm{d}y}{\mathrm{d}x}&=2\alpha\left( \frac{1}{x} \right)\\\frac1y\dfrac{\mathrm{d}y}{\mathrm{d}x}&=\frac{2\ln(x)}{x}\\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=y\left( \frac{2\ln(x)}{x} \right)\\&=\frac{2x^{\ln(x)-1}\ln(x)}{x}\\\dfrac{\mathrm{d}y}{\mathrm{d}x} &=2x^{\ln(x)-1}\ln(x)\end{aligned}\tag*{11}$
I tried asking various sources including AI sites whether there is a match between the functions $e^{\ln^2(x)}$ and $x^{\ln(x)}$ because the derivatives are the same and I checked the results my way of doing this with the help of Integral Calculator and looks as follows.
$\displaystyle \int x^{\ln(x)-1}\ln(x)\,\mathrm dx$" />
$u= \ln(x)$)" />
$u= x^{\ln(x)}$) and the antiderivative results via maxima" />
I also prove that what I'm doing is correct by differentiating the function for the result of the integral in the first way as shown in (5) using logarithmic differentiation as I did in (6) and how I get $x^{\ln (x)}$ from $e^{\ln^2(x)}$ I separate the $\ln^2(x)$ into $\ln(x)\ln(x)$ then because I use exponential rule that $\left(\alpha^{m}\right)^n =\alpha^{mn}$ and $e^{\ln(x)}= x$ because $e^{\ln(\alpha )}=1^{\alpha}=\alpha$. So do you think what I'm doing is right?
It looks like you're correct. From the properties of logarithms you have $a^b=\exp(b\ln a)$, so $x^{\ln x}=\exp[(\ln x)(\ln x)]=\exp[(\ln x)^2]$ as claimed.