What is the integral of 1/(64e^{-8x)+e^{8x}} ?

The integral of 1/(64e^{-8x)+e^{8x}} is -1/64 arctan(8e^{-8x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(64e^{-8x)+e^{8x}}

\int \frac{1}{64 e ^{- 8 x} + e ^{8 x}} d x

- \frac{1}{64} arctan (8 e^{- 8 x}) + C

Solution steps

\int \frac{1}{64 e ^{- 8 x} + e ^{8 x}} d x

Apply u-substitution

: \int - \frac{1}{8 ( 64 u ^{2} + 1 )} d u

= \int - \frac{1}{8 ( 64 u ^{2} + 1 )} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{64 u ^{2} + 1} d u

Apply Integral Substitution

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

Take the constant out:

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

= - \frac{1}{8} \cdot \frac{1}{8} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= - \frac{1}{8} \cdot \frac{1}{8} arctan (v)

Substitute back

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

Simplify

- \frac{1}{8} \cdot \frac{1}{8} arctan (8 e^{- 8 x}) : - \frac{1}{64} arctan (8 e^{- 8 x})

= - \frac{1}{64} arctan (8 e^{- 8 x})

Add a constant to the solution

= - \frac{1}{64} arctan (8 e^{- 8 x}) + C

(5, 7), (- 4, - 2)

\int (e^{4 x} + e^{- 4 x})^{2} d x

f (x) = - \frac{10}{x ^{2}}

\int 8 - 2 x^{2} d x

\frac{\partial}{\partial y} (4 arctan (\frac{x}{y}))

What is the integral of 1/(64e^{-8x)+e^{8x}} ?
The integral of 1/(64e^{-8x)+e^{8x}} is -1/64 arctan(8e^{-8x})+C