Question

If \(b=\int_{0}^{1} \frac{e^{t}}{t+1} d t\), then \(\int_{a-1}^{a} \frac{e^{-1}}{t-a-1}\) is
(A) be
(B) be
(C) \(-\mathrm{be}^{-\mathrm{a}}\)
(D) \(-\mathrm{be}^{\mathrm{a}}\)

Answer 1

Ans: (C)
Hint \(: \int_{a-1}^{a} \frac{e^{-t}}{t-a-1} d t\)
$$
\begin{aligned}
&=\int_{a-1}^{a} \frac{e^{-(2 a-1-t)}}{2 a-1-t-a-1} d t \\
&=\int_{a-1}^{a} \frac{e^{t-2 a+1}}{a-2-t} d t \\
&t-(a-1)=x \quad \Rightarrow d t=d x \\
&=\int_{0}^{1} \frac{e^{x} \cdot e^{-a}}{-x-1} d x \\
&=-e^{-a} \int_{0}^{1} \frac{e^{x}}{x+1} d x=-b e^{-a}
\end{aligned}
$$