The integral I = $\int \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{3\log_e x} - e^{2\log_e x}} dx$ is equal to

The integral $I = \int \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{3\log_e x} - e^{2\log_e x}} dx$ can be simplified using the property $e^{a\log_e x} = x^a$.

Applying this property to each term:

$e^{5\log_e x} = x^5$

$e^{4\log_e x} = x^4$

$e^{3\log_e x} = x^3$

$e^{2\log_e x} = x^2$

The integral becomes:

$I = \int \frac{x^5 - x^4}{x^3 - x^2} dx$

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Factoring the numerator and denominator:

Numerator: $x^5 - x^4 = x^4(x - 1)$

Denominator: $x^3 - x^2 = x^2(x - 1)$

$I = \int \frac{x^4(x - 1)}{x^2(x - 1)} dx$

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Canceling $(x-1)$ from numerator and denominator, and simplifying:

$I = \int \frac{x^4}{x^2} dx$

$I = \int x^2 dx$

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Using the power rule for integration:

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

Therefore, the integral equals $\frac{x^3}{3} + C$.