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

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

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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: \[ I = -\ln|\cos(1 + \log x)| + C \]
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