$\begin{aligned} & \int x_{\text {II }}^4(\log x)^3 d x \\ = & (\log x)^3 \cdot \int x^4 d x-\int\left(3(\log x)^2 \cdot \frac{1}{x} \cdot \int x^4 d x\right) d x \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{5} \int(\log x)^2 x^4 d x \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{5}\left[(\log x)^2 \cdot \frac{x^5}{5}-\int 2(\log x) \cdot \frac{1}{x} \cdot \frac{x^5}{5} d x\right] \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{5}\left[\frac{x^5}{5}(\log x)^2-\frac{2}{5} \int(\log x) \cdot x^4 d x\right] \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{25} x^5(\log x)^2+\frac{6}{25}\left[\frac{x^5}{5} \log x-\int \frac{1}{x} \cdot \frac{x^5}{5} d x\right] \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{25} x^5(\log x)^2+\frac{6}{125} x^5 \log x-\frac{6 x^5}{625}+K \\ = & x^5\left[\frac{1}{5}(\log x)^3-\frac{3}{25}(\log x)^2+\frac{6}{125} \log x-\frac{6}{625}\right]+K \\ A & \frac{1}{5}, B=\frac{-3}{25}, C=\frac{6}{125}, D=\frac{-6}{625} \\ A+B+C+5 D=\frac{1}{5}-\frac{3}{25}+\frac{6}{125}-\frac{6}{125} & \\ = & \frac{25-15+6-6}{125}=\frac{10}{125}=\frac{2}{25} .\end{aligned}$