Question:
$\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x$
Solution:
Let $\mathrm{I}=\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^{4}}-e^{\log x^{3}}}{e^{\log x^{6}}-e^{\log x^{9}}} d x$
$=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x$ (Using properties of logarithms)
$=\int \frac{x^{2}\left(x^{4}-x^{3}\right)}{x^{4}-x^{3}} d x=\int x^{2} d x=\frac{1}{3} x^{3}+C$
Therefore,
$\mathrm{I}=\int x^{2} d x=\frac{1}{3} x^{3}+\mathrm{C}$