Let `A = int_(0)^(1)(e^(t))/(1+t) dt`, then `int_(a-1)^(a)(e^(-1))/(t-a-1) dt` has the value :

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_{A-1}^{A} \frac{e^{-t}}{t - (A - 1)} \, dt \]

where \( A = \int_{0}^{1} \frac{e^{t}}{1+t} \, dt \).

### Step 1: Evaluate \( A \)
First, we need to compute the value of \( A \):

\[ A = \int_{0}^{1} \frac{e^{t}}{1+t} \, dt \]

This integral does not have a simple closed form, but we can denote its value as \( A \).

### Step 2: Change of Variables
Next, we will perform a change of variables in the integral \( I \). Let's set:

\[ x = t - (A - 1) \implies t = x + (A - 1) \]

Then, the differential \( dt = dx \). When \( t = A - 1 \), \( x = 0 \) and when \( t = A \), \( x = 1 \). Thus, we can rewrite the integral \( I \):

\[ I = \int_{0}^{1} \frac{e^{-(x + (A - 1))}}{x} \, dx \]

### Step 3: Simplify the Integral
Now, we simplify the expression inside the integral:

\[ I = \int_{0}^{1} \frac{e^{-(A - 1)}}{x} e^{-x} \, dx \]

This can be factored out:

\[ I = e^{-(A - 1)} \int_{0}^{1} \frac{e^{-x}}{x} \, dx \]

### Step 4: Evaluate the Integral
The integral \( \int_{0}^{1} \frac{e^{-x}}{x} \, dx \) diverges, but we can use the known result of the integral:

\[ \int_{0}^{1} e^{-x} \, dx = 1 - e^{-1} \]

However, we need to consider the limit as \( x \to 0 \). The integral diverges logarithmically.

### Step 5: Final Calculation
Thus, we can conclude that:

\[ I = e^{-(A - 1)} \cdot \text{(diverges)} \]

Therefore, the value of the integral \( \int_{A-1}^{A} \frac{e^{-t}}{t - (A - 1)} \, dt \) diverges.

### Final Answer
The value of the integral diverges.