Answer
${{x}^{n+1}}\left[ \frac{\ln \left| x \right|}{n+1}-\frac{1}{{{\left( n+1 \right)}^{2}}} \right]+C,\text{ }n\ne -1$
Work Step by Step
\[\begin{align}
& \int{{{x}^{n}}\ln \left| x \right|}dx \\
& \text{Integrate by parts} \\
& \text{Let }u=\ln \left| x \right|\to du=\frac{1}{x}dx \\
& dv={{x}^{n}}dx\to v=\frac{{{x}^{n+1}}}{n+1} \\
& \text{Using the integration by parts formula} \\
& \int{udv}=uv-\int{vdu} \\
& \int{{{x}^{n}}\ln \left| x \right|}dx=\frac{{{x}^{n+1}}}{n+1}\ln \left| x \right|-\int{\left( \frac{{{x}^{n+1}}}{n+1} \right)\left( \frac{1}{x} \right)dx} \\
& \int{{{x}^{n}}\ln \left| x \right|}dx=\frac{{{x}^{n+1}}}{n+1}\ln \left| x \right|-\frac{1}{n+1}\int{{{x}^{n}}dx} \\
& \int{{{x}^{n}}\ln \left| x \right|}dx=\frac{{{x}^{n+1}}}{n+1}\ln \left| x \right|-\frac{1}{n+1}\left( \frac{{{x}^{n+1}}}{n+1} \right)+C \\
& \text{Factoring out }{{x}^{n+1}} \\
& \int{{{x}^{n}}\ln \left| x \right|}dx={{x}^{n+1}}\left[ \frac{\ln \left| x \right|}{n+1}-\frac{1}{{{\left( n+1 \right)}^{2}}} \right]+C,\text{ }n\ne -1 \\
\end{align}\]
