Let A=∫ limits01 (et d t/1+t) then ∫ limitsa-1a (e-t d t/t-a-1) has the value

Question

Let A=∫ limits01 (et d t/1+t) then ∫ limitsa-1a (e-t d t/t-a-1) has the value (A) Ae -a (B) - Ae - a  (C) - Aae -1 (D) Ae a

Tardigrade Admin · Accepted Answer

I=∫ limits a -1 a ( e - t / t - a -1) dt put t = a -1+ y (so that lower limit becomes zero) ∴ I =∫ limits01 ( e 1- a - y / y -2) dy (now usingking) I=∫ limits01 (e1-a-1+y/1-y-2) d y=-e-a ∫ limits01 (ey/1+y) d y=-e-a A ⇒(B)

Q. Let $A = 0 \int 1 \frac{e ^{t} d t}{1 + t}$ then $a - 1 \int a \frac{e ^{- t} d t}{t - a - 1}$ has the value

$A e^{- a}$

$- A e^{- a}$

$- A a e^{- 1}$

$A e^{a}$

Q. Let A=0∫1​1+tetdt​ then a−1∫a​t−a−1e−tdt​ has the value

Ae−a

−Ae−a

−Aae−1

Aea

I=a−1∫a​t−a−1e−t​dt put t=a−1+y (so that lower limit becomes zero) ∴I=0∫1​y−2e1−a−y​dy (now usingking) I=0∫1​1−y−2e1−a−1+y​dy=−e−a0∫1​1+yey​dy=−e−aA⇒(B)

Q. Let $A = 0 \int 1 \frac{e ^{t} d t}{1 + t}$ then $a - 1 \int a \frac{e ^{- t} d t}{t - a - 1}$ has the value

$A e^{- a}$

$- A e^{- a}$

$- A a e^{- 1}$

$A e^{a}$