For $x > 1$, $\int \frac{e^{7\log x} - e^{5\log x}}{e^{5\log x} - e^{4\log x}} dx$ equals.

For $x > 1$, the integral can be simplified using the property $e^{a\log x} = x^a$.

Applying this property:

$e^{7\log x} = x^7$

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

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

The integral becomes:

$\int \frac{x^7 - x^5}{x^5 - x^4} dx$

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Factoring the numerator and denominator:

$x^7 - x^5 = x^5(x^2 - 1)$

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

The integral becomes:

$\int \frac{x^5(x^2 - 1)}{x^4(x - 1)} dx$

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

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The numerator can be factored as a difference of squares:

$x^2 - 1 = (x+1)(x-1)$

Substituting:

$\int \frac{x(x+1)(x-1)}{x-1} dx$

Since $x > 1$, $(x-1) \neq 0$:

$= \int x(x+1) dx$

$= \int (x^2 + x) dx$

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Integrating term by term:

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

Therefore, the integral equals $\dfrac{x^3}{3} + \dfrac{x^2}{2} + C$ where $C$ is a constant of integration.