Evaluate $\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} dx$.

$\frac{x^3}{3} + C$

The correct answer is Option (2) → $\frac{x^3}{3} + C$

$I = \int \left( \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} \right) dx$

$= \int \left( \frac{e^{\log x^6} - e^{\log x^5}}{e^{\log x^4} - e^{\log x^3}} \right) dx \quad [∵a \log b = \log b^a]$

$= \int \left( \frac{x^6 - x^5}{x^4 - x^3} \right) dx \quad [∵e^{\log x} = x]$

$= \int \frac{x^5(x - 1)}{x^3(x - 1)} dx$

$= \int x^2 dx = \frac{x^3}{3} + C$