What is the integral of 1/(64e^{-5x)+e^{5x}} ?

The integral of 1/(64e^{-5x)+e^{5x}} is -1/40 arctan(8e^{-5x})+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^{-5x)+e^{5x}}

\int \frac{1}{64 e ^{- 5 x} + e ^{5 x}} d x

- \frac{1}{40} arctan (8 e^{- 5 x}) + C

Solution steps

\int \frac{1}{64 e ^{- 5 x} + e ^{5 x}} d x

Apply u-substitution

: \int - \frac{1}{5 ( 64 u ^{2} + 1 )} d u

= \int - \frac{1}{5 ( 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}{5} \cdot \int \frac{1}{64 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{5} \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}{5} \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}{5} \cdot \frac{1}{8} arctan (v)

Substitute back

= - \frac{1}{5} \cdot \frac{1}{8} arctan (8 e^{- 5 x})

Simplify

- \frac{1}{5} \cdot \frac{1}{8} arctan (8 e^{- 5 x}) : - \frac{1}{40} arctan (8 e^{- 5 x})

= - \frac{1}{40} arctan (8 e^{- 5 x})

Add a constant to the solution

= - \frac{1}{40} arctan (8 e^{- 5 x}) + C

θ \to 0 lim (cot (π θ))

x \to 8 lim (\frac{f ( x )}{g ( x )})

\frac{d}{d x} (\frac{4}{1 - 9 x})

- \frac{x}{3}

\frac{d}{d x} (- \frac{y}{x ^{2} + y ^{2}})

What is the integral of 1/(64e^{-5x)+e^{5x}} ?
The integral of 1/(64e^{-5x)+e^{5x}} is -1/40 arctan(8e^{-5x})+C