What is the integral of 1/(49e^{-4x)+e^{4x}} ?

The integral of 1/(49e^{-4x)+e^{4x}} is -1/28 arctan(7e^{-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/(49e^{-4x)+e^{4x}}

\int \frac{1}{49 e ^{- 4 x} + e ^{4 x}} d x

- \frac{1}{28} arctan (7 e^{- 4 x}) + C

Solution steps

\int \frac{1}{49 e ^{- 4 x} + e ^{4 x}} d x

Apply u-substitution

: \int - \frac{1}{4 ( 49 u ^{2} + 1 )} d u

= \int - \frac{1}{4 ( 49 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}{49 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{4} \cdot \int \frac{1}{7 ( 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}{7} \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}{7} arctan (v)

Substitute back

= - \frac{1}{4} \cdot \frac{1}{7} arctan (7 e^{- 4 x})

Simplify

- \frac{1}{4} \cdot \frac{1}{7} arctan (7 e^{- 4 x}) : - \frac{1}{28} arctan (7 e^{- 4 x})

= - \frac{1}{28} arctan (7 e^{- 4 x})

Add a constant to the solution

= - \frac{1}{28} arctan (7 e^{- 4 x}) + C

ln (1 + 4 x), 0

\frac{25}{s ^{2} + 6 s + 25}

\int arctan (8 t) d t

(3 x y + 3 y - 4) d x + (x + 1)^{2} d y = 0

n = 0 \sum \infty n e^{- 8 n}

What is the integral of 1/(49e^{-4x)+e^{4x}} ?
The integral of 1/(49e^{-4x)+e^{4x}} is -1/28 arctan(7e^{-4x})+C