What is the integral of (e^{-2x})/((1+e^{-x))^3} ?

The integral of (e^{-2x})/((1+e^{-x))^3} is 1/(1+e^{-x)}-1/(2(1+e^{-x))^2}+C

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integral of (e^{-2x})/((1+e^{-x))^3}

\int \frac{e ^{- 2 x}}{( 1 + e ^{- x} ) ^{3}} d x

\frac{1}{1 + e ^{- x}} - \frac{1}{2 ( 1 + e ^{- x} ) ^{2}} + C

Solution steps

\int \frac{e ^{- 2 x}}{( 1 + e ^{- x} ) ^{3}} d x

Apply u-substitution

: \int - \frac{u}{( 1 + u ) ^{3}} d u

= \int - \frac{u}{( 1 + u ) ^{3}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{u}{( 1 + u ) ^{3}} d u

Apply u-substitution

: \int \frac{v - 1}{v ^{3}} d v

= - \int \frac{v - 1}{v ^{3}} d v

Expand

\frac{v - 1}{v ^{3}} : \frac{1}{v ^{2}} - \frac{1}{v ^{3}}

= - \int \frac{1}{v ^{2}} - \frac{1}{v ^{3}} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{1}{v ^{2}} d v - \int \frac{1}{v ^{3}} d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

= - (- \frac{1}{v} - (- \frac{1}{2 v ^{2}}))

Substitute back

= - (- \frac{1}{1 + e ^{- x}} - (- \frac{1}{2 ( 1 + e ^{- x} ) ^{2}}))

Simplify

= \frac{1}{1 + e ^{- x}} - \frac{1}{2 ( 1 + e ^{- x} ) ^{2}}

Add a constant to the solution

= \frac{1}{1 + e ^{- x}} - \frac{1}{2 ( 1 + e ^{- x} ) ^{2}} + C

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What is the integral of (e^{-2x})/((1+e^{-x))^3} ?
The integral of (e^{-2x})/((1+e^{-x))^3} is 1/(1+e^{-x)}-1/(2(1+e^{-x))^2}+C