Answer
$\dfrac{1}{2} (\ln x)^2 +C$
Work Step by Step
We will use integration by parts to obtain:
$\int x^n \ln x \ dx =\ln x \int x^{n} \ dx - \int ( \dfrac{d}{dx} (\ln x ) \int x^n \ dx) dx \\= \dfrac{x^{n+1}}{n+1} \ln x - \dfrac{1}{n+1} \int x^n dx \\= \dfrac{x^{n+1}}{n+1} (\ln x - \dfrac{1}{n+1} )+C$
Now, we will apply the substitution method. Let us consider that $\ln x=a \implies da=x^{-1} dx$ for $n=-1$
$=\int a da \\=\dfrac{a^2}{2}+C \\ =\dfrac{1}{2} (\ln x)^2 +C$
