The value of the integral `int_(1)^(2)e^(x)(log_(e)x+(x+1)/(x))dx` is

Options

A. `e^(2)(1+log_(e)2)`
B. `e^(2)-e`
C. `e^(2)(1+log_(e)2)-e`
D. `e^(2)-e(1+log_(e)2)`

Answer

To solve the integral

\[ I = \int_{1}^{2} e^x \left( \log_e x + \frac{x+1}{x} \right) dx, \]

we can break it down step-by-step.

### Step 1: Simplify the integrand
We can rewrite the integrand:

\[ \frac{x + 1}{x} = 1 + \frac{1}{x}. \]

Thus, the integral becomes:

\[ I = \int_{1}^{2} e^x \left( \log_e x + 1 + \frac{1}{x} \right) dx. \]

### Step 2: Split the integral
We can split the integral into three parts:

\[ I = \int_{1}^{2} e^x \log_e x \, dx + \int_{1}^{2} e^x \, dx + \int_{1}^{2} e^x \frac{1}{x} \, dx. \]

### Step 3: Evaluate the second integral
The second integral is straightforward:

\[ \int_{1}^{2} e^x \, dx = \left[ e^x \right]_{1}^{2} = e^2 - e. \]

### Step 4: Evaluate the first integral using integration by parts
For the first integral, we will use integration by parts. Let:

- \( u = \log_e x \) \(\Rightarrow du = \frac{1}{x} dx\)
- \( dv = e^x dx \) \(\Rightarrow v = e^x\)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ \int e^x \log_e x \, dx = e^x \log_e x - \int e^x \cdot \frac{1}{x} \, dx. \]

### Step 5: Substitute back into the integral
Now substituting back into the integral \( I \):

\[ I = \left[ e^x \log_e x \right]_{1}^{2} - \int_{1}^{2} e^x \frac{1}{x} \, dx + (e^2 - e). \]

### Step 6: Evaluate the boundary terms
Now we evaluate the boundary terms:

\[ \left[ e^x \log_e x \right]_{1}^{2} = e^2 \log_e 2 - e^1 \log_e 1 = e^2 \log_e 2 - 0 = e^2 \log_e 2. \]

### Step 7: Combine all parts
Now we have:

\[ I = e^2 \log_e 2 - \int_{1}^{2} e^x \frac{1}{x} \, dx + (e^2 - e). \]

### Step 8: Recognize the integral
Notice that the integral \( \int_{1}^{2} e^x \frac{1}{x} \, dx \) is the same as the term we derived from integration by parts. Thus, we can simplify:

\[ I = e^2 \log_e 2 + e^2 - e - \int_{1}^{2} e^x \frac{1}{x} \, dx. \]

### Step 9: Final evaluation
Combining all the terms, we get:

\[ I = e^2 (1 + \log_e 2) - e. \]

### Final Answer
Thus, the value of the integral is:

\[ I = e^2 (1 + \log_e 2) - e. \]