(i) `int (log x)/(x(1+log x)(2+log x)) dx`

Answer

To solve the integral \(\int \frac{\log x}{x(1+\log x)(2+\log x)} \, dx\), we can use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( u = \log x \). Then, the differential \( du = \frac{1}{x} \, dx \) implies that \( dx = x \, du = e^u \, du \).

### Step 2: Rewrite the Integral
Substituting \( u \) into the integral, we have:
\[ \int \frac{u}{e^u(1+u)(2+u)} e^u \, du = \int \frac{u}{(1+u)(2+u)} \, du \]

### Step 3: Partial Fraction Decomposition
Next, we can perform partial fraction decomposition on \(\frac{u}{(1+u)(2+u)}\):
\[ \frac{u}{(1+u)(2+u)} = \frac{A}{1+u} + \frac{B}{2+u} \]
Multiplying through by the denominator \((1+u)(2+u)\) gives:
\[ u = A(2+u) + B(1+u) \]
Expanding this, we have:
\[ u = 2A + Au + B + Bu \]
Combining like terms:
\[ u = (A + B)u + (2A + B) \]
Setting coefficients equal gives us the system:
1. \( A + B = 1 \)
2. \( 2A + B = 0 \)

### Step 4: Solve for A and B
From the first equation, \( B = 1 - A \). Substituting into the second equation:
\[ 2A + (1 - A) = 0 \implies A + 1 = 0 \implies A = -1 \]
Then substituting back to find \( B \):
\[ B = 1 - (-1) = 2 \]

### Step 5: Rewrite the Integral
Thus, we can rewrite the integral as:
\[ \int \left( \frac{-1}{1+u} + \frac{2}{2+u} \right) \, du \]

### Step 6: Integrate
Now we can integrate each term:
\[ \int \frac{-1}{1+u} \, du + 2 \int \frac{1}{2+u} \, du = -\log|1+u| + 2\log|2+u| + C \]

### Step 7: Substitute Back
Substituting back \( u = \log x \):
\[ -\log|1+\log x| + 2\log|2+\log x| + C \]

### Final Answer
Thus, the final answer is:
\[ 2\log|2+\log x| - \log|1+\log x| + C \]