What is the integral of 1/(9e^{-8x)+e^{8x}} ?

The integral of 1/(9e^{-8x)+e^{8x}} is -1/24 arctan(3e^{-8x})+C

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integral of 1/(9e^{-8x)+e^{8x}}

\int \frac{1}{9 e ^{- 8 x} + e ^{8 x}} d x

- \frac{1}{24} arctan (3 e^{- 8 x}) + C

Solution steps

\int \frac{1}{9 e ^{- 8 x} + e ^{8 x}} d x

Apply u-substitution

: \int - \frac{1}{8 ( 9 u ^{2} + 1 )} d u

= \int - \frac{1}{8 ( 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}{8} \cdot \int \frac{1}{9 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{8} \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}{8} \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}{8} \cdot \frac{1}{3} arctan (v)

Substitute back

= - \frac{1}{8} \cdot \frac{1}{3} arctan (3 e^{- 8 x})

Simplify

- \frac{1}{8} \cdot \frac{1}{3} arctan (3 e^{- 8 x}) : - \frac{1}{24} arctan (3 e^{- 8 x})

= - \frac{1}{24} arctan (3 e^{- 8 x})

Add a constant to the solution

= - \frac{1}{24} arctan (3 e^{- 8 x}) + C

f (x) = x^{2} x - 5 x

\frac{d}{d x} (\frac{1 - 5 x}{3 + x})

\frac{\partial}{\partial x} (x arctan (\frac{y}{z}))

f (r) = e^{r} + x^{e}

0.1 x^{3} - 0.1 x^{2} + 50 x

What is the integral of 1/(9e^{-8x)+e^{8x}} ?
The integral of 1/(9e^{-8x)+e^{8x}} is -1/24 arctan(3e^{-8x})+C