Let, the function F be defined as F x = ∫ 1 x e t t d t ,   x > 0 , then the value of the…

Given $F\left(x\right)={\int }_1^x\frac{e^t}{t}dt$

Let, $I={\int }_1^x\frac{e^t}{t+a}dt$

Let, $t+a=y$   $\Rightarrow dt=dy$

Also, $t=1 \Rightarrow y=1+a$ and $t=x \Rightarrow y=x+a$

$\therefore I={\int }_{1+a}^{x+a}\frac{e^{y-a}}{y}dy$

$\Rightarrow I=e^{-a}{\int }_{1+a}^{x+a}\frac{e^y}{y}dy$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(t\right)dt,$ we get

$I=e^{-a}{\int }_{1+a}^{x+a}\frac{e^t}{t}dt$

$\Rightarrow I=e^{-a}\left[{\int }_1^{1+a}\frac{e^t}{t}dt+{\int }_{1+a}^{x+a}\frac{e^t}{t}dt-{\int }_1^{1+a}\frac{e^t}{t}dt\right]$

Using ${\int }_a^{b}f\left(x\right)dx+{\int }_b^{c}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx, a<c<b,$

$\Rightarrow I=e^{-a}\left[{\int }_1^{x+a}\frac{e^t}{t}dt-{\int }_1^{1+a}\frac{e^t}{t}dt\right]$

Using, the given relation, we get

$I=e^{-a}\left[F\left(x+a\right)-F\left(1+a\right)\right].$