Evaluate: `int(1)/(x)ln((x)/(e^(x)))dx=`

Options

A. `(1)/(2)e^(x)-lnx+C`
B. `(1)/(2)lnx-e^(x)+C`
C. `(1)/(2)(lnx)^(2)-x+C`
D. `(e^(x))/(2x)+C`

Explanation

To evaluate the integral \( I = \int \frac{1}{x} \ln\left(\frac{x}{e^x}\right) \, dx \), we can follow these steps:

### Step 1: Simplify the logarithm
Using the property of logarithms, we can rewrite the integrand:
\[ \ln\left(\frac{x}{e^x}\right) = \ln(x) - \ln(e^x) = \ln(x) - x \]
Thus, we can express the integral as:
\[ I = \int \frac{1}{x} \left(\ln(x) - x\right) \, dx \]

### Step 2: Split the integral
We can separate the integral into two parts:
\[ I = \int \frac{\ln(x)}{x} \, dx - \int 1 \, dx \]

### Step 3: Evaluate the first integral
The first integral, \( \int \frac{\ln(x)}{x} \, dx \), can be solved using substitution. Let \( t = \ln(x) \), then \( dt = \frac{1}{x} \, dx \) or \( dx = x \, dt = e^t \, dt \). Therefore:
\[ \int \frac{\ln(x)}{x} \, dx = \int t \, dt = \frac{t^2}{2} + C = \frac{(\ln(x))^2}{2} + C \]

### Step 4: Evaluate the second integral
The second integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 5: Combine the results
Putting it all together, we have:
\[ I = \frac{(\ln(x))^2}{2} - x + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{1}{x} \ln\left(\frac{x}{e^x}\right) \, dx = \frac{(\ln(x))^2}{2} - x + C \]

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