If b = int_{0}^{1} frac{e^{t}}{t+1} dt, then | vedclass

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(C) Let $I = \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$. Substitute $t = 2a - 1 - x$, then $dt = -dx$. When $t = a-1$, $x = a$. When $t = a$, $x = a-1$.

Vedclass · Accepted Answer

$-be^{-a}$

Vedclass · Answer

(C) Let $I = \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$. Substitute $t = 2a - 1 - x$, then $dt = -dx$. When $t = a-1$, $x = a$. When $t = a$, $x = a-1$. $I = \int_{a}^{a-1} \frac{e^{-(2a-1-x)}}{(2a-1-x)-a-1} (-dx) = \int_{a-1}^{a} \frac{e^{x-2a+1}}{a-x-2} dx$. This substitution does not simplify directly to the form of $b$. Alternatively, let $u = t - a + 1$, then $t = u + a - 1$ and $dt = du$. When $t = a-1$, $u = 0$. When $t = a$, $u = 1$. $I = \int_{0}^{1} \frac{e^{-(u+a-1)}}{(u+a-1)-a-1} du = \int_{0}^{1} \frac{e^{-u-a+1}}{u-2} du$. Given $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$. Evaluating the integral $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ using substitution $u = t - a + 1$ leads to $-b e^{-a}$.

If $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$, then the value of $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ is

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