`int_(1)^(2) (dx)/(x(1+log x)^(2))`

Answer

To solve the integral

\[ \int_{1}^{2} \frac{dx}{x(1 + \log x)^{2}}, \]

we will follow these steps:

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

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

Since \( x = e^{t-1} \) (from \( t = 1 + \log x \)), we can express \( dx \) in terms of \( t \):

\[ dx = e^{t-1} dt. \]

### Step 2: Change of Limits
Next, we need to change the limits of integration.

- When \( x = 1 \):
\[ t = 1 + \log 1 = 1 + 0 = 1. \]

- When \( x = 2 \):
\[ t = 1 + \log 2. \]

Thus, the new limits are from \( t = 1 \) to \( t = 1 + \log 2 \).

### Step 3: Rewrite the Integral
Now we can rewrite the integral:

\[ \int_{1}^{2} \frac{dx}{x(1 + \log x)^{2}} = \int_{1}^{1 + \log 2} \frac{1}{t^{2}} dt. \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{1}{t^{2}} dt \) is:

\[ \int \frac{1}{t^{2}} dt = -\frac{1}{t}. \]

Now we evaluate this from \( t = 1 \) to \( t = 1 + \log 2 \):

\[ \left[-\frac{1}{t}\right]_{1}^{1 + \log 2} = -\frac{1}{1 + \log 2} + \frac{1}{1} = 1 - \frac{1}{1 + \log 2}. \]

### Step 5: Simplify the Result
Now we simplify:

\[ 1 - \frac{1}{1 + \log 2} = \frac{(1 + \log 2) - 1}{1 + \log 2} = \frac{\log 2}{1 + \log 2}. \]

Thus, the value of the definite integral is:

\[ \int_{1}^{2} \frac{dx}{x(1 + \log x)^{2}} = \frac{\log 2}{1 + \log 2}. \]

### Final Answer
\[ \int_{1}^{2} \frac{dx}{x(1 + \log x)^{2}} = \frac{\log 2}{1 + \log 2}. \]

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