Evaluate `int(1+x^(2)log_(e)x)/(x+x^(2)log_(e)x)dx`

Explanation

To evaluate the integral

\[ I = \int \frac{1 + x^2 \log_e x}{x + x^2 \log_e x} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Factor out \(x\) from the denominator

We can rewrite the denominator:

\[ x + x^2 \log_e x = x(1 + x \log_e x). \]

Thus, we can express the integral as:

\[ I = \int \frac{1 + x^2 \log_e x}{x(1 + x \log_e x)} \, dx. \]

### Step 2: Split the fraction

Now we can split the fraction:

\[ I = \int \left( \frac{1}{x} + \frac{x \log_e x}{1 + x \log_e x} \right) \, dx. \]

### Step 3: Separate the integral

This gives us two separate integrals:

\[ I = \int \frac{1}{x} \, dx + \int \frac{x \log_e x}{1 + x \log_e x} \, dx. \]

### Step 4: Evaluate the first integral

The first integral is straightforward:

\[ \int \frac{1}{x} \, dx = \log_e |x| + C_1. \]

### Step 5: Evaluate the second integral

For the second integral, we will use substitution. Let:

\[ t = 1 + x \log_e x. \]

Now we differentiate \(t\):

\[ dt = \left( \log_e x + 1 \right) dx. \]

From this, we can express \(dx\):

\[ dx = \frac{dt}{\log_e x + 1}. \]

Now we need to express \(\log_e x\) in terms of \(t\):

\[ \log_e x = \frac{t - 1}{x}. \]

### Step 6: Substitute back into the integral

Substituting \(dx\) and \(\log_e x\) into the second integral:

\[ \int \frac{x \log_e x}{1 + x \log_e x} \, dx = \int \frac{x \frac{t - 1}{x}}{t} \cdot \frac{dt}{\log_e x + 1} = \int \frac{t - 1}{t} \cdot \frac{dt}{\log_e x + 1}. \]

### Step 7: Simplify and integrate

This integral can be split further, and we can integrate each part separately.

The final result will combine the results from both integrals:

\[ I = \log_e |x| + \text{(result from the second integral)} + C. \]

### Final Answer

Thus, the final answer can be expressed as:

\[ I = \log_e |x| + \text{(result from the second integral)} + C. \]