Answer
The solution is $$\frac{x^6}{6}(\ln x-\frac{1}{6})+c.$$
Work Step by Step
To solve the integral
$$\int x^5\ln xdx$$ we will integrate by parts using formula $\int udv=uv-\int vdu$ and for $u$ and $v$ we will use $u=\ln x$ and $dv=x^5dx$ which gives us $du=\frac{dx}{x}$ and $v=\frac{x^6}{6}$. Putting all this into our integral we have:
$$\int x^5\ln xdx=\int udv=uv-\int vdu=\frac{1}{6}x^6\ln x-\int\frac{1}{6}x^6\frac{dx}{x}=\frac{1}{6}x^6\ln x-\frac{1}{6}\int x^5dx=\int\frac{1}{6}x^6\ln x-\frac{1}{6}\frac{x^6}{6}+c=\frac{x^6}{6}(\ln x-\frac{1}{6})+c,$$
where $c$ is arbitrary constant.
