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

If 01etdtt+1=a then, b1betdttb1 is equal to

  1. A

    aeb

  2. B

    -aeb

  3. C

    bea

  4. D

    aeb

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    Solution:

    b1bettb1dt=10e(t+b)(t+b)b1dt=eb10ett1dt=eb01e(t1)t2dt

    Put,

    t1=sdt=ds=eb10es(s+1)ds=eb10ess+1ds=eb01ett+1dt=aeb

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