Let the function F be defined as F(x) = int_{ | vedclass

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(D) Given $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt$ for $x > 0$.Let $I = \int_{1}^{x} \frac{e^{t}}{t+a} dt$.Substitute $t+a = z$,so $t = z-a$ and $dt =

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$e^{-a} [F(x+a) - F(1+a)]$

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(D) Given $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt$ for $x > 0$.Let $I = \int_{1}^{x} \frac{e^{t}}{t+a} dt$.Substitute $t+a = z$,so $t = z-a$ and $dt = dz$.When $t = 1$,$z = 1+a$. When $t = x$,$z = x+a$.Substituting these into the integral:$I = \int_{1+a}^{x+a} \frac{e^{z-a}}{z} dz = e^{-a} \int_{1+a}^{x+a} \frac{e^{z}}{z} dz$.Using the property of definite integrals $\int_{b}^{c} f(z) dz = \int_{1}^{c} f(z) dz - \int_{1}^{b} f(z) dz$:$I = e^{-a} \left[ \int_{1}^{x+a} \frac{e^{z}}{z} dz - \int_{1}^{1+a} \frac{e^{z}}{z} dz \right]$.By the definition of $F(x)$,this becomes:$I = e^{-a} [F(x+a) - F(1+a)]$.Thus,the correct option is $D$.

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

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