Question:
$\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$
Solution:
$\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}=\frac{e^{4 \log x}\left(e^{\log x}-1\right)}{e^{2 \log x}\left(e^{\log x}-1\right)}$
$=e^{2 \log x}$
$=e^{\log x^{2}}$
$=x^{2}$
$\therefore \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int x^{2} d x=\frac{x^{3}}{3}+\mathrm{C}$