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Ifint0^1(e^t dt)/(t+1)=a ,t h e ne v a l...

`Ifint_0^1(e^t dt)/(t+1)=a ,t h e ne v a l u a t eint_(b-1)^b(e^(-t)dt)/(t-b-1)`

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Verified by Experts

The correct Answer is:
`-ae^(-b)`
In `int_(b-1)^(b)(e^(-t)dt)/(t-b-1)` but `t-b-1=-y-1`. So `dt=-dy`
`:.int_(b-1)^(b)(e^(-t)dt)/(t-b-1)=int_(1)^(0)(e^(y-b))/(-y-1)(-dy)`
`=-e^(-b)int_(0)^(1)(e^(y))/(y+1)dy=-ae^(-b)`
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