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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Factor

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

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

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

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

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 2 u + 1 )} d u

= \int \frac{1}{2 ( u ^{2} + 2 u + 1 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 2 u + 1} d u

Complete the square

u^{2} + 2 u + 1 : (u + 1)^{2}

= \frac{1}{2} \cdot \int \frac{1}{( u + 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

f (x) = 4 - x, x = - 2, x = 3, y = 0

\frac{\partial}{\partial x} (s^{2} + t^{2})

t^{3} e^{- 2 t} - 3 cos (5 t)

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