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

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

Popular >

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

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

\frac{1}{1 + e ^{- x}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= - (- \frac{1}{u})

Substitute back

u = 1 + e^{- x}

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

Simplify

= \frac{1}{1 + e ^{- x}}

Add a constant to the solution

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

\int (\frac{1}{sin ^{2} ( x ) cos ^{2} ( x )}) d x

\int \frac{x cos ( x )}{( sin ( x ) ) ^{3}} d x

\int x^{3} + 7 d x

\int \frac{( x ^{2} )}{4} d x

\int \frac{12 x ^{11}}{x ^{12} + 1} d x

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