`int(tan(1+log x))/(x) dx`

Answer

To solve the integral \( I = \int \frac{\tan(1 + \log x)}{x} \, dx \), we will use substitution and integration techniques.

### Step-by-Step Solution:

1. **Substitution**: Let \( t = 1 + \log x \). This means that \( \log x = t - 1 \).

2. **Differentiate**: Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(1 + \log x) = \frac{dt}{dx} \]
The derivative of \( 1 \) is \( 0 \) and the derivative of \( \log x \) (base \( e \)) is \( \frac{1}{x} \). Thus,
\[ \frac{dt}{dx} = \frac{1}{x} \]
Rearranging gives:
\[ dx = x \, dt \]

3. **Substituting in the Integral**: Since \( \log x = t - 1 \), we can express \( x \) in terms of \( t \):
\[ x = e^{t-1} \]
Therefore, \( dx = e^{t-1} \, dt \).

4. **Rewrite the Integral**: Substitute \( t \) and \( dx \) into the integral:
\[ I = \int \tan(t) \cdot \frac{e^{t-1}}{e^{t-1}} \, dt = \int \tan(t) \, dt \]

5. **Integrate**: The integral of \( \tan(t) \) is:
\[ \int \tan(t) \, dt = -\ln|\cos(t)| + C \]

6. **Back Substitute**: Recall that \( t = 1 + \log x \):
\[ I = -\ln|\cos(1 + \log x)| + C \]

### Final Answer:
Thus, the evaluated integral is:
\[ I = -\ln|\cos(1 + \log x)| + C \]