Answer
The result is $$\int_{1}^{2}x^4\ln xdx=\frac{32}{5}\ln2-\frac{31}{25}.$$
Work Step by Step
To solve the integral $$\int_{1}^{2}x^4\ln xdx$$ we will integrate by parts using that $u=\ln x$ and $x^4dx=dv$ which gives us $du=\frac{dx}{x}$ and $v=\frac{x^5}{5}$. Putting this into our integral we have:
$$\int_{1}^{2}x^4\ln xdx=\left.\frac{x^5}{5}\ln x\right|_{1}^{2}-\int_{1}^{2}\frac{x^5}{5}\frac{dx}{x}=\frac{32}{5}\ln2-\frac{1}{5}\ln1-\frac{1}{5}\int_{1}^{2}x^4dx=\frac{32}{5}\ln2-\frac{1}{5}\left.\frac{x^5}{5}\right|_{1}^{2}=\frac{32}{5}\ln2-\frac{1}{25}(32-1)=\frac{32}{5}\ln2-\frac{31}{25}.$$
