The integral $$\int \frac{e^{3\log_e 2x} + 5e^{2\log_e 2x}}{e^{4\log_e x} + 5e^{3\log_e x} - 7e^{2\log_e x}} dx$$, $$x > 0$$, is equal to (where $$c$$ is a constant of integration)

We simplify the integrand using the identity $$e^{k\log_e m} = m^k$$.

In the numerator: $$e^{3\log_e 2x} = (2x)^3 = 8x^3$$ and $$e^{2\log_e 2x} = (2x)^2 = 4x^2$$. So the numerator is $$8x^3 + 5 \cdot 4x^2 = 8x^3 + 20x^2$$.

In the denominator: $$e^{4\log_e x} = x^4$$, $$e^{3\log_e x} = x^3$$, and $$e^{2\log_e x} = x^2$$. So the denominator is $$x^4 + 5x^3 - 7x^2$$.

The integral becomes $$\int \frac{8x^3 + 20x^2}{x^4 + 5x^3 - 7x^2}\,dx = \int \frac{8x^3 + 20x^2}{x^2(x^2 + 5x - 7)}\,dx = \int \frac{8x + 20}{x^2 + 5x - 7}\,dx$$.

We note that $$\frac{d}{dx}(x^2 + 5x - 7) = 2x + 5$$, and $$8x + 20 = 4(2x + 5)$$.

Therefore, the integral is $$4\int \frac{2x + 5}{x^2 + 5x - 7}\,dx = 4\log_e|x^2 + 5x - 7| + c$$.