Let `f(x)=int_(0)^(x)(e^(t))/(t)dt(xgt0),`
then `e^(-a)[f(x+1)-f(1+a)]=`

Options

A. `int_(0)^(x)=(e^(t))/((t+a))dt`
B. `int_(1)^(x)(e^(t))/(t+a)dt`
C. `e^(-a)int_(1+a)^(x+a)(e^(t))/(t)dt`
D. `int_(0)^(x)(e^(t-a))/((t+a))dt`

Explanation

To solve the problem, we need to evaluate the expression \( e^{-a}[f(x+1)-f(1+a)] \), where \( f(x) = \int_{0}^{x} \frac{e^t}{t} dt \).

### Step-by-Step Solution:

1. **Define the function \( f(x) \)**:
\[ f(x) = \int_{0}^{x} \frac{e^t}{t} dt \]

2. **Substitute \( x+1 \) and \( 1+a \) into \( f \)**:
\[ f(x+1) = \int_{0}^{x+1} \frac{e^t}{t} dt \]
\[ f(1+a) = \int_{0}^{1+a} \frac{e^t}{t} dt \]

3. **Calculate \( f(x+1) - f(1+a) \)**:
\[ f(x+1) - f(1+a) = \int_{0}^{x+1} \frac{e^t}{t} dt - \int_{0}^{1+a} \frac{e^t}{t} dt \]
This can be rewritten using the property of integrals:
\[ = \int_{1+a}^{x+1} \frac{e^t}{t} dt \]

4. **Multiply by \( e^{-a} \)**:
\[ e^{-a}[f(x+1) - f(1+a)] = e^{-a} \int_{1+a}^{x+1} \frac{e^t}{t} dt \]

5. **Evaluate the integral**:
To evaluate the integral \( \int_{1+a}^{x+1} \frac{e^t}{t} dt \), we can leave it in this form as it represents the area under the curve \( \frac{e^t}{t} \) from \( t = 1+a \) to \( t = x+1 \).

6. **Final expression**:
Thus, the final expression is:
\[ e^{-a} \int_{1+a}^{x+1} \frac{e^t}{t} dt \]

### Final Answer:
\[ e^{-a}[f(x+1) - f(1+a)] = e^{-a} \int_{1+a}^{x+1} \frac{e^t}{t} dt \]