What is the integral of+1/(25e^{-4x)+e^{4x}} ?

The integral of+1/(25e^{-4x)+e^{4x}} is -1/20 arctan(5e^{-4x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of+1/(25e^{-4x)+e^{4x}}

\int + \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

- \frac{1}{20} arctan (5 e^{- 4 x}) + C

Solution steps

\int \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

Apply u-substitution

: \int - \frac{1}{4 ( 25 u ^{2} + 1 )} d u

= \int - \frac{1}{4 ( 25 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}{4} \cdot \int \frac{1}{25 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{4} \cdot \int \frac{1}{5 ( 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}{4} \cdot \frac{1}{5} \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}{4} \cdot \frac{1}{5} arctan (v)

Substitute back

= - \frac{1}{4} \cdot \frac{1}{5} arctan (5 e^{- 4 x})

Simplify

- \frac{1}{4} \cdot \frac{1}{5} arctan (5 e^{- 4 x}) : - \frac{1}{20} arctan (5 e^{- 4 x})

= - \frac{1}{20} arctan (5 e^{- 4 x})

Add a constant to the solution

= - \frac{1}{20} arctan (5 e^{- 4 x}) + C

\frac{d}{d θ} (cot^{2} (sin (θ)))

\int \frac{x + 8}{x} d x

\frac{d}{d t} (e^{4 t} cos (t))

\int y^{- 2} d y

\frac{d}{d x} (\frac{7 x - 4}{- 3 x + 3})

What is the integral of+1/(25e^{-4x)+e^{4x}} ?
The integral of+1/(25e^{-4x)+e^{4x}} is -1/20 arctan(5e^{-4x})+C