`int (dx)/(x(1+log_(e)x)(3+log_(e)x))`

Answer

To solve the integral

\[ I = \int \frac{dx}{x(1 + \log_e x)(3 + \log_e x)}, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = \log_e x \). Then, we differentiate both sides:

\[ dt = \frac{1}{x} dx \quad \Rightarrow \quad dx = x dt = e^t dt. \]

Substituting \( x = e^t \) into the integral, we have:

\[ I = \int \frac{e^t dt}{e^t(1 + t)(3 + t)} = \int \frac{dt}{(1 + t)(3 + t)}. \]

### Step 2: Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on the integrand:

\[ \frac{1}{(1 + t)(3 + t)} = \frac{A}{1 + t} + \frac{B}{3 + t}. \]

Multiplying through by the denominator \((1 + t)(3 + t)\), we get:

\[ 1 = A(3 + t) + B(1 + t). \]

Expanding this gives:

\[ 1 = 3A + At + B + Bt = (A + B)t + (3A + B). \]

### Step 3: Setting Up the System of Equations
Now, we can set up the system of equations by equating coefficients:

1. \( A + B = 0 \) (coefficient of \( t \))
2. \( 3A + B = 1 \) (constant term)

### Step 4: Solving the System
From the first equation, we can express \( B \) in terms of \( A \):

\[ B = -A. \]

Substituting \( B = -A \) into the second equation:

\[ 3A - A = 1 \quad \Rightarrow \quad 2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}. \]

Then, substituting back to find \( B \):

\[ B = -\frac{1}{2}. \]

### Step 5: Writing the Integral
Now we can rewrite the integral using the values of \( A \) and \( B \):

\[ I = \int \left( \frac{1/2}{1 + t} - \frac{1/2}{3 + t} \right) dt = \frac{1}{2} \int \frac{dt}{1 + t} - \frac{1}{2} \int \frac{dt}{3 + t}. \]

### Step 6: Integrating
Now we can integrate each term:

\[ I = \frac{1}{2} \log |1 + t| - \frac{1}{2} \log |3 + t| + C, \]

where \( C \) is the constant of integration.

### Step 7: Simplifying
Using the properties of logarithms, we can combine the logarithms:

\[ I = \frac{1}{2} \log \left| \frac{1 + t}{3 + t} \right| + C. \]

### Step 8: Back Substitution
Finally, we substitute back \( t = \log_e x \):

\[ I = \frac{1}{2} \log \left| \frac{1 + \log_e x}{3 + \log_e x} \right| + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ \int \frac{dx}{x(1 + \log_e x)(3 + \log_e x)} = \frac{1}{2} \log \left| \frac{1 + \log_e x}{3 + \log_e x} \right| + C. \]