`int log_e xdx = int 1/(log_x e) dx = `

Answer

To solve the integral \( \int \log_e x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \log_e x \, dx \]
Using the change of base formula for logarithms, we can express this as:
\[ I = \int \frac{1}{\log_x e} \, dx \]
However, we will proceed with the first form for simplicity.

### Step 2: Use Integration by Parts
We will apply integration by parts, where we let:
- \( u = \log_e x \) (thus \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (thus \( v = x \))

According to the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ I = x \log_e x - \int x \cdot \frac{1}{x} \, dx \]

### Step 3: Simplify the Integral
The integral simplifies to:
\[ I = x \log_e x - \int 1 \, dx \]
Calculating the remaining integral gives:
\[ I = x \log_e x - x + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \log_e x \, dx = x \log_e x - x + C \]