What is the integral of 1/(81e^{-5x)+e^{5x}} ?

The integral of 1/(81e^{-5x)+e^{5x}} is -1/45 arctan(9e^{-5x})+C

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integral of 1/(81e^{-5x)+e^{5x}}

\int \frac{1}{81 e ^{- 5 x} + e ^{5 x}} d x

- \frac{1}{45} arctan (9 e^{- 5 x}) + C

Solution steps

\int \frac{1}{81 e ^{- 5 x} + e ^{5 x}} d x

Apply u-substitution

: \int - \frac{1}{5 ( 81 u ^{2} + 1 )} d u

= \int - \frac{1}{5 ( 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}{5} \cdot \int \frac{1}{81 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{5} \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}{5} \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}{5} \cdot \frac{1}{9} arctan (v)

Substitute back

= - \frac{1}{5} \cdot \frac{1}{9} arctan (9 e^{- 5 x})

Simplify

- \frac{1}{5} \cdot \frac{1}{9} arctan (9 e^{- 5 x}) : - \frac{1}{45} arctan (9 e^{- 5 x})

= - \frac{1}{45} arctan (9 e^{- 5 x})

Add a constant to the solution

= - \frac{1}{45} arctan (9 e^{- 5 x}) + C

\int_{0}^{5} y + 4 d y

y^{^{'}} - (\frac{7}{8}) y = - \frac{7}{8} e^{- 7 x} y^{9}

2 c

\frac{d y}{d x} = \frac{5 x \cdot sec ( \frac{y}{x} ) + y}{x}

\frac{d y}{d x} + 12 y = 14

What is the integral of 1/(81e^{-5x)+e^{5x}} ?
The integral of 1/(81e^{-5x)+e^{5x}} is -1/45 arctan(9e^{-5x})+C