`int(x^(e-1)-e^(x-1))/(x^e-e^x)dx`

Answer

To solve the integral \( \int \frac{x^{e-1} - e^{x-1}}{x^e - e^x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral for clarity:
\[ \int \frac{x^{e-1} - e^{x-1}}{x^e - e^x} \, dx = \int \frac{x^{e-1} - \frac{e^x}{e}}{x^e - e^x} \, dx \]

### Step 2: Simplify the numerator
We can factor out \( \frac{1}{e} \) from the numerator:
\[ = \int \frac{1}{e} \cdot \frac{e x^{e-1} - e^x}{x^e - e^x} \, dx \]

### Step 3: Recognize the derivative
Notice that the numerator \( e x^{e-1} - e^x \) is the derivative of \( x^e - e^x \). Thus, we can set:
\[ t = x^e - e^x \]
Then, the derivative \( dt \) is given by:
\[ dt = (e x^{e-1} - e^x) \, dx \]

### Step 4: Substitute and change variables
Now we can substitute \( dt \) into our integral:
\[ \int \frac{1}{e} \cdot \frac{dt}{t} \]

### Step 5: Integrate
The integral of \( \frac{1}{t} \) is:
\[ \frac{1}{e} \ln |t| + C \]

### Step 6: Substitute back for \( t \)
Substituting back for \( t \):
\[ = \frac{1}{e} \ln |x^e - e^x| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{x^{e-1} - e^{x-1}}{x^e - e^x} \, dx = \frac{1}{e} \ln |x^e - e^x| + C \]