If B=∫1/(e^x+1)dx=-f(x)+C, where, C is the constant of integration and e^f0=2, then the value of e^(f-1) is

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MARKS Content Team · Accepted Answer

Correct answer: (B) e+1. The given integral is B=∫(e^(-x))/(1+e^(-x))dx Let, e^(-x)+1=t ⇒-e^(-x)dx=dt ∴B=∫(-dt)/t=-ln⁡t+C ⇒B=-ln⁡e^(-x)+1+C ∴fx=ln⁡e^(-x)+1 Thus, e^(f-1)=e^(ln⁡e+1)=e+1

If $B = \int \frac{1}{e^{x} + 1} d x$ $= - f (x) + C$ , where, $C$ is the constant of integration and $e^{f (0)} = 2,$ then the value of $e^{f (- 1)}$ is

Solution

More Indefinite Integration Questions

If B=∫1ex+1dx=-f(x)+C, where, C is the constant of integration and ef0=2, then the value of ef-1 is

Solution

More Indefinite Integration Questions

If $B = \int \frac{1}{e^{x} + 1} d x$ $= - f (x) + C$ , where, $C$ is the constant of integration and $e^{f (0)} = 2,$ then the value of $e^{f (- 1)}$ is