Let, the function F be defined as $$F(x) = \int_1^x \frac{e^t}{t} dt$$, $$x > 0$$, then the value of the integral $$\int_1^x \frac{e^t}{t+a} dt$$, where a > 0, is:

We are given the function $$ F(x) = \int_1^x \frac{e^t}{t} \, dt $$ for $$ x > 0 $$, and we need to evaluate the integral $$ \int_1^x \frac{e^t}{t+a} \, dt $$ where $$ a > 0 $$.

Notice that the integrand $$ \frac{e^t}{t+a} $$ resembles the integrand in $$ F(x) $$, but with $$ t $$ shifted by $$ a $$ in the denominator. To simplify, we use the substitution $$ u = t + a $$. Then, $$ du = dt $$, and when $$ t = 1 $$, $$ u = 1 + a $$, and when $$ t = x $$, $$ u = x + a $$. Substituting these into the integral:

$$ \int_1^x \frac{e^t}{t+a} \, dt = \int_{1+a}^{x+a} \frac{e^{u - a}}{u} \, du $$

Since $$ e^{u - a} = e^u \cdot e^{-a} $$, we can rewrite the integral as:

$$ \int_{1+a}^{x+a} \frac{e^u \cdot e^{-a}}{u} \, du = e^{-a} \int_{1+a}^{x+a} \frac{e^u}{u} \, du $$

The integral $$ \int_{1+a}^{x+a} \frac{e^u}{u} \, du $$ can be expressed in terms of the function $$ F $$. Recall that $$ F(z) = \int_1^z \frac{e^t}{t} \, dt $$. Changing the variable from $$ t $$ to $$ u $$ for clarity, $$ F(u) = \int_1^u \frac{e^s}{s} \, ds $$. Therefore:

$$ \int_{1+a}^{x+a} \frac{e^u}{u} \, du = F(x+a) - F(1+a) $$

This is because $$ F(x+a) = \int_1^{x+a} \frac{e^s}{s} \, ds $$ and $$ F(1+a) = \int_1^{1+a} \frac{e^s}{s} \, ds $$, so subtracting them gives the integral from $$ 1+a $$ to $$ x+a $$.

Substituting back, we get:

$$ e^{-a} \int_{1+a}^{x+a} \frac{e^u}{u} \, du = e^{-a} \left[ F(x+a) - F(1+a) \right] $$

Thus, the value of the integral $$ \int_1^x \frac{e^t}{t+a} \, dt $$ is $$ e^{-a} \left[ F(x+a) - F(1+a) \right] $$.

Comparing with the options:

• Option A: $$ e^a [F(x) - F(1+a)] $$
• Option B: $$ e^{-a} [F(x+a) - F(a)] $$
• Option C: $$ e^a [F(x+a) - F(1+a)] $$
• Option D: $$ e^{-a} [F(x+a) - F(1+a)] $$

Our result matches option D exactly.

Hence, the correct answer is Option D.