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

Explanation

To evaluate the integral \( \int \frac{dx}{x + x \log x} \), we can follow these steps:

### Step 1: Simplify the Denominator
We start with the integral:
\[ \int \frac{dx}{x + x \log x} \]
We can factor out \( x \) from the denominator:
\[ = \int \frac{dx}{x(1 + \log x)} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral as:
\[ = \int \frac{1}{x(1 + \log x)} \, dx \]

### Step 3: Substitution
We will use the substitution:
\[ t = 1 + \log x \]
Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \frac{1}{x} \quad \Rightarrow \quad dt = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dt \]
Since \( x = e^{t-1} \) (from \( \log x = t - 1 \)), we can substitute \( x \) back into the expression for \( dx \):
\[ dx = e^{t-1} \, dt \]

### Step 4: Substitute in the Integral
Now we substitute \( dx \) and \( x \) into the integral:
\[ = \int \frac{e^{t-1} \, dt}{e^{t-1}(1 + \log(e^{t-1}))} \]
Since \( \log(e^{t-1}) = t - 1 \), we have:
\[ = \int \frac{e^{t-1} \, dt}{e^{t-1}(t)} = \int \frac{dt}{t} \]

### Step 5: Integrate
Now we can integrate:
\[ = \log |t| + C \]

### Step 6: Substitute Back
Finally, we substitute back \( t = 1 + \log x \):
\[ = \log |1 + \log x| + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{dx}{x + x \log x} = \log |1 + \log x| + C \]