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

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Correct option is B

∫b−1b e−tt−b−1dt=∫−10 e−(t+b)(t+b)−b−1dt=e−b∫−10 e−tt−1dt=e−b∫01 e−(t−1)t−2dtPut,t−1=−s⇒dt=−ds=−e−b∫10 es−(s+1)ds=e−b∫10 ess+1ds=−e−b∫01 ett+1dt=−ae−b

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If ∫01 etdtt+1=a then, ∫b−1b e−tdtt−b−1 is equal to

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