$\mathrm{The}\mathrm{integral}I=\int \frac{e^{5\log ex}-e^{4\log ex}}{e^{3\log ex}-e^{2\log ex}}\mathrm{ d}x\mathrm{is}\mathrm{equal}\mathrm{to}$

The correct answer is:
$\frac{x^3}{3}+C,\text{where C is the constant of integration}$

Step-by-Step Explanation:

We are given the following integral to evaluate:
$I=\int \frac{e^{5{\log }_{e}x}-e^{4{\log }_{e}x}}{e^{3{\log }_{e}x}-e^{2{\log }_{e}x}}dx$

To simplify this expression, we use a fundamental property of logarithms and exponents:
$a\cdot {\log }_{e}b={\log }_e\left(b^a\right)$

Applying this property to each term in the exponents:
$5{\log }_{e}x={\log }_e\left(x^5\right)$
$4{\log }_{e}x={\log }_e\left(x^4\right)$
$3{\log }_{e}x={\log }_e\left(x^3\right)$
$2{\log }_{e}x={\log }_e\left(x^2\right)$

Next, we use another key identity relating exponents and logarithms with the same base:
$e^{{\log }_{e}y}=y$

Substituting this identity back into our terms:
$e^{5{\log }_{e}x}=e^{{\log }_e\left(x^5\right)}=x^5$
Similarly:
$e^{4{\log }_{e}x}=x^4$
$e^{3{\log }_{e}x}=x^3$
$e^{2{\log }_{e}x}=x^2$

Substitute these simplified expressions back into the integral:
$I=\int \frac{x^5-x^4}{x^3-x^2}dx$

We can factor out common terms from the numerator and denominator:
$I=\int \frac{x^4(x-1)}{x^2(x-1)}dx$

Assuming $x>0$ (so that the logarithms are defined) and $x\ne 1$, we cancel the common factor $(x-1)$ and simplify the fraction:
$I=\int \frac{x^4}{x^2}dx=\int x^{2}dx$

Now, apply the power rule for integration, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$:
$I=\frac{x^3}{3}+C$