How do you integrate #int e^-x/(1+e^-x)dx#?
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The answer is
Perform the substitution
Therefore,
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Let (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method.To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u =To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (uTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + eTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u =To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( 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^{-xTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( 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}),To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). ThenTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), thenTo integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then,To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to (x), you can use the substitution method. Let (u = 1 + e^{-x}), then (To integrate ( \frac{e^{-x}}{1+e^{-x}} ) with respect to ( x ), you can use the substitution method. Let ( u = 1 + e^{-x} ). Then, ( 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 =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 = -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 = -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^{-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 = -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^{-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^{-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}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}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} 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\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 \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).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 ).
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), weTo 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 (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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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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