What is the integral of 1/(9e^{-6x)+e^{6x}} ?

The integral of 1/(9e^{-6x)+e^{6x}} is -1/18 arctan(3e^{-6x})+C

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integral of 1/(9e^{-6x)+e^{6x}}

\int \frac{1}{9 e ^{- 6 x} + e ^{6 x}} d x

- \frac{1}{18} arctan (3 e^{- 6 x}) + C

Solution steps

\int \frac{1}{9 e ^{- 6 x} + e ^{6 x}} d x

Apply u-substitution

= \int - \frac{1}{6 ( 9 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}{9 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{6} \cdot \int \frac{1}{3 ( 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}{3} \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}{3} arctan (v)

Substitute back

= - \frac{1}{6} \cdot \frac{1}{3} arctan (3 e^{- 6 x})

Simplify

- \frac{1}{6} \cdot \frac{1}{3} arctan (3 e^{- 6 x}) : - \frac{1}{18} arctan (3 e^{- 6 x})

= - \frac{1}{18} arctan (3 e^{- 6 x})

Add a constant to the solution

= - \frac{1}{18} arctan (3 e^{- 6 x}) + C

3 y^{^{''}} + 27 y = 0

d er i v a t i v e \frac{32}{x}

\frac{\partial}{\partial x} (e^{x} - e^{x^{2}})

\frac{\partial}{\partial x} (600 - 4 x + 10 x^{2} y)

d er i v a t i v e f (x) = \frac{x ^{2} - 5 x + 4}{x ^{2} + 5 x + 4}

What is the integral of 1/(9e^{-6x)+e^{6x}} ?
The integral of 1/(9e^{-6x)+e^{6x}} is -1/18 arctan(3e^{-6x})+C