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Question

Question: \[\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{1 +}\mathbf{e}^{\mathbf{x}}}\mathbf{=}}\]...

∫dx1+ex=\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{1 +}\mathbf{e}^{\mathbf{x}}}\mathbf{=}}

A

log⁡(1+ex)\log(1 + e^{x})

B

−log⁡(1+e−x)- \log(1 + e^{- x})

C

−log⁡(1−e−x)- \log(1 - e^{- x})

D

log⁡(e−x+e−2x)\log(e^{- x} + e^{- 2x})

Answer

−log⁡(1+e−x)- \log(1 + e^{- x})

Explanation

Solution

I=∫dx1+ex=∫e−xe−x+1dx\int_{}^{}{\frac{dx}{1 + e^{x}} = \int_{}^{}{\frac{e^{- x}}{e^{- x} + 1}dx}}, Put e−x+1=te^{- x} + 1 = t ⇒ −e−xdx=dt- e^{- x}dx = dt ⇒ I=−∫dtt=−log⁡t= - \int_{}^{}\frac{dt}{t} = - \log t =−log⁡(1+e−x)= - \log(1 + e^{- x})