If \( \int_{0}^{1} \frac{e^{t} d t}{t+1}=a \), then \( \int_{b-1}^{b} \frac{e^{-t} d t}{t-b-1} \...

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If \( \int_{0}^{1} \frac{e^{t} d t}{t+1}=a \), then \( \int_{b-1}^{b} \frac{e^{-t} d t}{t-b-1} \) is equal to
(A) \( a e^{-b} \)
(B) \( -a e^{-b} \)
(C) \( b e^{-b} \)
(D) None of these
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