What is the integral of-1/(e^x(8+e^{-x))^2} ?

The integral of-1/(e^x(8+e^{-x))^2} is -1/(8+e^{-x)}+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(8+e^{-x))^2}

\int - \frac{1}{e ^{x} ( 8 + e ^{- x} ) ^{2}} d x

- \frac{1}{8 + e ^{- x}} + C

Solution steps

\int - \frac{1}{e ^{x} ( 8 + e ^{- x} ) ^{2}} d x

Take the constant out:

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

= - \int \frac{1}{e ^{x} ( 8 + e ^{- x} ) ^{2}} d x

Apply u-substitution

: \int - \frac{1}{( 8 + u ) ^{2}} d u

= - \int - \frac{1}{( 8 + 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}{( 8 + u ) ^{2}} d u)

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

= - (- (- \frac{1}{8 + e ^{- x}}))

Simplify

= - \frac{1}{8 + e ^{- x}}

Add a constant to the solution

= - \frac{1}{8 + e ^{- x}} + C

x \to 0 + lim (1 4^{x} + 2)

e^{s i n^{2} (x)}

f (x) = 7 - \frac{16}{x}, at x = 4

y = x^{2^{x}}

(x^{5} + 3 y) d x - x d y = 0

What is the integral of-1/(e^x(8+e^{-x))^2} ?
The integral of-1/(e^x(8+e^{-x))^2} is -1/(8+e^{-x)}+C