Evaluate `int (dx)/(x+xlogx)`

Explanation

To evaluate the integral

\[ \int \frac{dx}{x + x \log x}, \]

we can start by simplifying the expression in the denominator.

### Step 1: Factor out \( x \) from the denominator
We can factor \( x \) out of the denominator:

\[ x + x \log x = x(1 + \log x). \]

Thus, we can rewrite the integral as:

\[ \int \frac{dx}{x(1 + \log x)}. \]

### Step 2: Rewrite the integral
Now, we can separate the terms:

\[ \int \frac{dx}{x(1 + \log x)} = \int \frac{1}{x} \cdot \frac{1}{1 + \log x} \, dx. \]

### Step 3: Substitute \( t = 1 + \log x \)
Let’s make the substitution \( t = 1 + \log x \). Then, we differentiate \( t \):

\[ dt = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dt. \]

From the substitution \( t = 1 + \log x \), we can express \( x \) in terms of \( t \):

\[ \log x = t - 1 \quad \Rightarrow \quad x = e^{t-1} = \frac{e^t}{e}. \]

### Step 4: Substitute in the integral
Now substituting \( x \) and \( dx \) into the integral:

\[ \int \frac{1}{x(1 + \log x)} \, dx = \int \frac{1}{\frac{e^t}{e} \cdot t} \cdot \frac{e^t}{e} \, dt = \int \frac{1}{t} \, dt. \]

### Step 5: Evaluate the integral
The integral \( \int \frac{1}{t} \, dt \) is a standard integral:

\[ \int \frac{1}{t} \, dt = \log |t| + C. \]

### Step 6: Substitute back for \( t \)
Now, we substitute back \( t = 1 + \log x \):

\[ \log |t| + C = \log |1 + \log x| + C. \]

### Final Answer
Thus, the final answer is:

\[ \int \frac{dx}{x + x \log x} = \log |1 + \log x| + C. \]

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