How do you integrate #int e^-x/(1+e^-x)dx#?

Answer 1

#I=int(e^-x)/(1+e^-x)dx=int((-1)d/dx(1+e^-x))/(1+e^-x)=-ln|1+e^-x|#
#I=ln|1/(1+e^-x)|+c#

#I=int(e^-x)/(1+e^-x)dx#, take, #e^-x=t=>e^-x(-1)dx=dt=>e^-xdx=-dt# #I=-intdt/(1+t)=-ln|1+t|+c=-ln|1+e^-x|+c# #I=ln|1/(1+e^-x)|+c#
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Answer 2

# ln|1/(1+e^-x)|+C.#

Subst. #1+e^-x=t," so that, "-e^-xdx=dt#.
#:. I=inte^-x/(1+e^-x)dx#,
#=int-1/tdt=-int1/tdt#,
#=-ln|t|#,
#=ln|t|^-1#,
# rArr I=ln|1/(1+e^-x)|+C.#
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Answer 3

The answer is #=-ln(1+e^-x)+C#

Perform the substitution

#u=1+e^-x#, #=>#, #du=-e^-xdx#

Therefore,

#int(e^-xdx)/(1+e^-x)=-int(du)/(u)#
#=-lnu#
#=-ln(1+e^-x)+C#
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Answer 4

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Let (u = 1 + e^{-x}), then (du = -e^{-x}dx).To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

NowTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). RearTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now,To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). RearrangingTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substituteTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx),To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

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Now, substitute ( u ) and (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we getTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) intoTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dxTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integralTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\fracTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \intTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{duTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \fracTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{eTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-xTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-xTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}).To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx =To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). SubstitTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into theTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\intTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \intTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \fracTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \fracTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{eTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{uTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-xTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u}To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} duTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du =To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\lnTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-xTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|uTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute backTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dxTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( uTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx =To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\intTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 +To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + eTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \fracTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-xTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x}To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{uTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} )To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) intoTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}duTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into theTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the resultTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du \To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

ThisTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifiesTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\lnTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\ln|To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

[ -\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\ln|1To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

[ -\lnTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\ln|1+To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

[ -\ln|To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\ln|1+eTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

[ -\ln|u| + C = -\To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( du = -e^{-x} dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}} dx = -\int \frac{1}{u} du = -\ln|u| + C ]

Finally, substitute back ( u = 1 + e^{-x} ) into the result:

[ -\ln|1+e^{-x}| + C ]To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (du = -e^{-x}dx). Rearranging for (dx), we get (dx = -\frac{du}{e^{-x}}). Substituting these into the integral:

[ \int \frac{e^{-x}}{1+e^{-x}}dx = -\int \frac{1}{u}du ]

This simplifies to:

[ -\ln|u| + C = -\ln|1+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.

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.

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.

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.

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