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How do you evaluate the integral of #int e^-x/(1+e^-2x) dx#?

Answer 1

Scarlett Allison

#int (e^(-x)dx)/(1+e^(-2x))=-arctan e^(-x)+C#

#int (e^(-x)dx)/(1+e^(-2x))=-tan^(-1) e^(-x)+C#

#int e^(-x)/(1+e^(-2x))dx#

#int e^(-x)/(1^2+(e^(-x))^2)dx#

#-int (e^(-x)(-1)*dx)/(1^2+(e^(-x))^2)#

We can now use the formula

#int (du)/(u^2+a^2)=1/a*arctan (u/a)+C#

#int e^(-x)/(1+e^(-2x))dx=-int (e^(-x)(-1)*dx)/(1^2+(e^(-x))^2)#

#=(-1/1)*arctan(e^(-x)/1)+C#

#=-arctan e^(-x)+C#

#=-tan^(-1) e^(-x)+C#

God bless....I hope the explanation is useful.

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Answer 2

Aurora Alvarado

To evaluate the integral (\int \frac{e^{-x}}{1+e^{-2x}} , dx), we can use a substitution method. Let (u = e^{-x}), then (du = -e^{-x} , dx). Substituting these into the integral:

[ \begin{aligned} \int \frac{e^{-x}}{1+e^{-2x}} , dx &= \int \frac{1}{1+u^2} , (-du) \ &= -\int \frac{1}{1+u^2} , du \ &= -\arctan(u) + C \ &= -\arctan(e^{-x}) + C \end{aligned} ]

So, the integral evaluates to (-\arctan(e^{-x}) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.