What is the integral of 1/(81e^{-6x)+e^{6x}} ?

The integral of 1/(81e^{-6x)+e^{6x}} is -1/54 arctan(9e^{-6x})+C

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integral of 1/(81e^{-6x)+e^{6x}}

\int \frac{1}{81 e ^{- 6 x} + e ^{6 x}} d x

- \frac{1}{54} arctan (9 e^{- 6 x}) + C

Solution steps

\int \frac{1}{81 e ^{- 6 x} + e ^{6 x}} d x

Apply u-substitution

: \int - \frac{1}{6 ( 81 u ^{2} + 1 )} d u

= \int - \frac{1}{6 ( 81 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}{6} \cdot \int \frac{1}{81 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{6} \cdot \int \frac{1}{9 ( 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}{6} \cdot \frac{1}{9} \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}{6} \cdot \frac{1}{9} arctan (v)

Substitute back

= - \frac{1}{6} \cdot \frac{1}{9} arctan (9 e^{- 6 x})

Simplify

- \frac{1}{6} \cdot \frac{1}{9} arctan (9 e^{- 6 x}) : - \frac{1}{54} arctan (9 e^{- 6 x})

= - \frac{1}{54} arctan (9 e^{- 6 x})

Add a constant to the solution

= - \frac{1}{54} arctan (9 e^{- 6 x}) + C

\int_{1}^{x} 1 + 9 t d t

n \to \infty lim (n ln (n))

(5 x) + (4 - x^{2})

\int x e^{8 x} d x

\frac{d}{d x} (8^{\frac{1}{3}})

What is the integral of 1/(81e^{-6x)+e^{6x}} ?
The integral of 1/(81e^{-6x)+e^{6x}} is -1/54 arctan(9e^{-6x})+C