Evaluate: `int(log(1+1/x))/(x(1+x))dx`

Explanation

To evaluate the integral

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

we can follow these steps:

### Step 1: Substitution
Let \( t = \log(1 + \frac{1}{x}) \).

To find \( dt \), we differentiate \( t \):
\[ t = \log(1 + \frac{1}{x}) \implies dt = \frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) \, dx = -\frac{1}{x^2 + x} \, dx. \]

### Step 2: Rearranging \( dx \)
From the expression for \( dt \), we can rearrange it to express \( dx \):
\[ dx = - (x^2 + x) \, dt. \]

### Step 3: Substitute in the Integral
Now, substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{t}{x(1+x)} \cdot - (x^2 + x) \, dt. \]
Notice that \( x(1+x) = x^2 + x \), so the integral simplifies to:
\[ -\int t \, dt. \]

### Step 4: Integrate
Now, we can integrate:
\[ -\int t \, dt = -\frac{t^2}{2} + C. \]

### Step 5: Substitute Back for \( t \)
Now, substitute back \( t = \log(1 + \frac{1}{x}) \):
\[ -\frac{1}{2} \left(\log(1 + \frac{1}{x})\right)^2 + C. \]

### Final Answer
Thus, the final answer is:
\[ -\frac{1}{2} \left(\log(1 + \frac{1}{x})\right)^2 + C. \]

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