Let A=∫01 etdt1+t then ∫a−1a e−tdtt−a−1=

I=∫a−1a e−tdtt−(a+1) Put t=y-a−1⇒t=a-1⇒y=0; t=a⇒y=1I=∫01 e−y−a+1dyy−2=e−a∫01 e−(y−1)(y−1)−1dy =e-a∫01e-(1-y-1)1-y-2dy=e-a∫01ey-y-1dy=e−a⋅A(−1)

Let A=∫01 etdt1+t then ∫a−1a e−tdtt−a−1=

I=∫a−1a e−tdtt−(a+1) Put t=y-a−1⇒t=a-1⇒y=0; t=a⇒y=1I=∫01 e−y−a+1dyy−2=e−a∫01 e−(y−1)(y−1)−1dy =e-a∫01e-(1-y-1)1-y-2dy=e-a∫01ey-y-1dy=e−a⋅A(−1)

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Let $A = \int_{0}^{1} \frac{e^{t} dt}{1 + t} then \int_{a - 1}^{a} \frac{e^{- t} dt}{t - a - 1} =$

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$A e^{- a}$

$- A e^{- a}$

$- a e^{- a}$

$A e^{a}$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int_{a - 1}^{a} \frac{e^{- t} dt}{t - (a + 1)} Put t = y - (a - 1) \Rightarrow t = a - 1 \Rightarrow y = 0; t = a \Rightarrow y = 1 \\ I = \int_{0}^{1} \frac{e^{- y - a + 1} dy}{y - 2} = e^{- a} \int_{0}^{1} \frac{e^{- (y - 1)}}{(y - 1) - 1} d y = e^{- a} \int_{0}^{1} \frac{e^{- (1 - y - 1)}}{1 - y - 2} d y = e^{- a} \int_{0}^{1} \frac{e^{y}}{- y - 1} d y = e^{- a} \cdot A (- 1) \end{array}$