Find the value of $$ \int \log _{e} x \cdot d x $$

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$$ x\ln{(x)}-x + C $$ Calculate the integral $$ \int \log _{e} x \cdot d x $$. Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: $$\int \ln{\left(xight)} dx$$ - step1: Prepare for integration by parts: $$\begin{align}&u=\ln{\left(xight)}\&dv=dx\end{align}$$ - step2: Calculate the derivative: $$\begin{align}&du=\frac{1}{x}dx\&dv=dx\end{align}$$ - step3: Evaluate the integral: $$\begin{align}&du=\frac{1}{x}dx\&v=x\end{align}$$ - step4: Substitute the values into formula: $$\ln{\left(xight)}	imes x-\int \frac{1}{x}	imes x dx$$ - step5: Calculate: $$x\ln{\left(xight)}-\int 1 dx$$ - step6: Evaluate the integral: $$x\ln{\left(xight)}-x$$ - step7: Add the constant of integral C: $$x\ln{\left(xight)}-x + C, C \in \mathbb{R}$$ The value of $$ \int \log _{e} x \cdot d x $$ is $$ x\ln{(x)}-x + C $$, where $$ C $$ is an arbitrary constant.

Find the value of \( \int \log _{e} x \cdot d x \)

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Find the value of \( \int \log _{e} x \cdot d x \)

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