If B = ∫ 1 e x + 1 d x = - f ( x ) + C , where, C is the constant of integration and e f 0 = 2 , then the value of e f - 1 is

Question

If B = ∫ 1 e x + 1 d x = - f ( x ) + C , where, C is the constant of integration and e f 0 = 2 , then the value of e f - 1 is, 4, e + 1, 2 e, 0

MARKS App · Accepted Answer

The given integral is B = ∫ e - x 1 + e - x d x Let, e - x + 1 = t ⇒ - e - x d x = d t ∴ B = ∫ - d t t = - ln ⁡ t + C ⇒ B = - ln ⁡ e - x + 1 + C ∴ f x = ln ⁡ e - x + 1 Thus, e f - 1 = e ln ⁡ e + 1 = e + 1

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