`int(e^(6logx)-e^(5logx))/(e^(4logx)-e^(3logx))dx`

Explanation

To solve the integral

\[ \int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Exponential Expressions
Using the property of exponents and logarithms, we can rewrite \( e^{m \log n} \) as \( n^m \). Therefore:

\[ e^{6 \log x} = x^6, \quad e^{5 \log x} = x^5, \quad e^{4 \log x} = x^4, \quad e^{3 \log x} = x^3. \]

Substituting these into the integral gives:

\[ \int \frac{x^6 - x^5}{x^4 - x^3} \, dx. \]

### Step 2: Factor the Numerator and Denominator
We can factor out common terms from the numerator and the denominator:

- In the numerator \( x^6 - x^5 \), we can factor out \( x^5 \):

\[ x^6 - x^5 = x^5(x - 1). \]

- In the denominator \( x^4 - x^3 \), we can factor out \( x^3 \):

\[ x^4 - x^3 = x^3(x - 1). \]

Now the integral becomes:

\[ \int \frac{x^5(x - 1)}{x^3(x - 1)} \, dx. \]

### Step 3: Cancel Common Factors
Since \( x - 1 \) is common in both the numerator and the denominator, we can cancel it (assuming \( x \neq 1 \)):

\[ \int \frac{x^5}{x^3} \, dx = \int x^{5 - 3} \, dx = \int x^2 \, dx. \]

### Step 4: Integrate
Now we can integrate \( x^2 \):

\[ \int x^2 \, dx = \frac{x^3}{3} + C, \]

where \( C \) is the constant of integration.

### Final Answer
Thus, the final solution to the integral is:

\[ \frac{x^3}{3} + C. \]

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