What is the integral of 1/(81e^{-7x)+e^{7x}} ?

The integral of 1/(81e^{-7x)+e^{7x}} is -1/63 arctan(9e^{-7x})+C

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integral of 1/(81e^{-7x)+e^{7x}}

\int \frac{1}{81 e ^{- 7 x} + e ^{7 x}} d x

- \frac{1}{63} arctan (9 e^{- 7 x}) + C

Solution steps

\int \frac{1}{81 e ^{- 7 x} + e ^{7 x}} d x

Apply u-substitution

: \int - \frac{1}{7 ( 81 u ^{2} + 1 )} d u

= \int - \frac{1}{7 ( 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}{7} \cdot \int \frac{1}{81 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{7} \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}{7} \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}{7} \cdot \frac{1}{9} arctan (v)

Substitute back

= - \frac{1}{7} \cdot \frac{1}{9} arctan (9 e^{- 7 x})

Simplify

- \frac{1}{7} \cdot \frac{1}{9} arctan (9 e^{- 7 x}) : - \frac{1}{63} arctan (9 e^{- 7 x})

= - \frac{1}{63} arctan (9 e^{- 7 x})

Add a constant to the solution

= - \frac{1}{63} arctan (9 e^{- 7 x}) + C

\int \frac{( x ^{2} - 4 )}{x ( x ^{2} + 4 )} d x

\frac{\partial}{\partial x} (y^{2} z)

\int_{0.6}^{1.5} \frac{48 + 8 y}{81} d y

x \to \infty + lim (65.4 - 65.4 e^{(- \frac{7 x}{2180})})

(- 2, - 4), (2, 5)

What is the integral of 1/(81e^{-7x)+e^{7x}} ?
The integral of 1/(81e^{-7x)+e^{7x}} is -1/63 arctan(9e^{-7x})+C