# Integrate the following functions with respect to x. start integral e raised to the power of 5 times the logarithm of x minus e raised to the power of 4 times the logarithm of x, all over e raised to the power of 3 times the logarithm of x times x minus e raised to the power of 2 times the logarithm of x times x dx.

## Question

Integrate the following functions w.r.t. x.

$$ \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x} x-e^{2 \log x}} x $$

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## Answer

To solve the integral $$ \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x} x - e^{2 \log x}} dx, $$ we will simplify the expression step by step.

First, we recognize that $ e^{a \log x} = x^a $. This allows us to rewrite the integral as follows: $$ \int \frac{e^{5 \log x} - e^{4 \log x}}{e^{3 \log x} - e^{2 \log x}} dx = \int \frac{e^{\log x^5} - e^{\log x^4}}{e^{\log x^3} - e^{\log x^2}} dx. $$

Simplifying further, it becomes: $$ \int \frac{x^5 - x^4}{x^3 - x^2} dx. $$

Next, factor out the common terms in the numerator and the denominator: $$ \int \frac{x^4 (x - 1)}{x^2 (x - 1)} dx. $$

We see that $ (x - 1) $ cancels out: $$ \int \frac{x^4}{x^2} dx = \int x^2 dx. $$

Now we integrate $ x^2 $ with respect to $ x $: $$ \int x^2 dx = \frac{1}{3} x^3 + C. $$

Thus, the solution to the integral is: $$ \boxed{\frac{1}{3} x^3 + C} $$

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