Let A = intlimits_0^1 frac{e^t}{1 + t} , dt. | vedclass

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(B) Let $I = \int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$. Substitute $t = a - 1 + y$,so $dt = dy$. When $t = a - 1$,$y = 0$. When $t = a$,$y

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$-Ae^{-a}$

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(B) Let $I = \int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$. Substitute $t = a - 1 + y$,so $dt = dy$. When $t = a - 1$,$y = 0$. When $t = a$,$y = 1$. Substituting these into the integral: $I = \int\limits_0^1 \frac{e^{-(a - 1 + y)}}{a - 1 + y - a - 1} \, dy = \int\limits_0^1 \frac{e^{1 - a - y}}{y - 2} \, dy$. This does not match the form of $A$ directly. Let us re-evaluate the substitution. Let $t = a - 1 + y$. The denominator is $t - a - 1 = (a - 1 + y) - a - 1 = y - 2$. Actually,let us use the substitution $t = a - 1 + y$ again. $I = \int\limits_0^1 \frac{e^{-(a - 1 + y)}}{y - 2} \, dy = e^{1-a} \int\limits_0^1 \frac{e^{-y}}{y - 2} \, dy$. Wait,the standard property $\int_0^1 f(t) dt = \int_0^1 f(1-t) dt$ is applicable. Let $t = a - 1 + y$. Then $I = \int_0^1 \frac{e^{-(a-1+y)}}{y-2} dy = e^{1-a} \int_0^1 \frac{e^{-y}}{y-2} dy$. Using $y 	o 1-y$: $I = e^{1-a} \int_0^1 \frac{e^{-(1-y)}}{1-y-2} dy = e^{1-a} \int_0^1 \frac{e^{-1+y}}{-1-y} dy = -e^{1-a} e^{-1} \int_0^1 \frac{e^y}{1+y} dy = -e^{-a} A$.

Let $A = \int\limits_0^1 \frac{e^t}{1 + t} \, dt$. Then $\int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$ has the value:

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