A function is formed by the four exponential terms and each exponent of the every exponential term is the product of a number and logarithmic of $x$ to base $e$. The four exponential logarithmic form terms formed the function in division form by the subtraction of two exponential terms in both numerator and denominator.
In this problem, the integration of the function is required to find to obtain the solution of this integral problem.
$\displaystyle \int \dfrac{e^{\displaystyle 5 \log_{e} x} -e^{\displaystyle 4\log_{e} x}}{e^{\displaystyle 3\log_{e} x} -e^{\displaystyle 2\log_{e} x}} dx$
Observe each exponential term having a logarithmic term as it exponent in both numerator and denominator. The base of the exponent and the base of the logarithmic term are same and it is Napier constant $e$.
Each term can be simplified by the fundamental logarithmic identity but a number is additionally multiplied to each logarithmic term at the power position. However, it can be overcome by applying the power rule of logarithms.
$=$ $\displaystyle \int \dfrac{e^{\displaystyle \log_{e} x^5} -e^{\displaystyle \log_{e} x^4}}{e^{\displaystyle \log_{e} x^3} -e^{\displaystyle \log_{e} x^2}} dx$
Now, each exponential logarithmic term can be simplified according to the fundamental logarithmic identity.
$=$ $\displaystyle \int \dfrac{x^5 -x^4}{x^3 -x^2} \,dx$
Now, simplify the expression by using the concept of exponents.
$=$ $\displaystyle \int \dfrac{x^4(x-1)}{x^2(x-1)} \,dx$
$=$ $\displaystyle \int \dfrac{x^4 \times (x-1)}{x^2 \times (x-1)} \,dx$
$=$ $\require{\cancel} \displaystyle \int \dfrac{\cancel{x^4} \times \cancel{(x-1)}}{\cancel{x^2} \times \cancel{(x-1)}} \,dx$
$=$ $\displaystyle \int x^2 \,dx$
Now, use the integral of x raised to the power of n formula to obtain the solution of this integration problem.
$= \dfrac{x^{2+1}}{2+1} + C$
$= \dfrac{x^3}{3} + C$
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