What is the integral of 1/(9e^{-2x)+e^{2x}} ?

The integral of 1/(9e^{-2x)+e^{2x}} is -1/6 arctan(3e^{-2x})+C

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integral of 1/(9e^{-2x)+e^{2x}}

\int \frac{1}{9 e ^{- 2 x} + e ^{2 x}} d x

- \frac{1}{6} arctan (3 e^{- 2 x}) + C

Solution steps

\int \frac{1}{9 e ^{- 2 x} + e ^{2 x}} d x

Apply u-substitution

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

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

Apply Integral Substitution

= - \frac{1}{2} \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}{2} \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}{2} \cdot \frac{1}{3} arctan (v)

Substitute back

= - \frac{1}{2} \cdot \frac{1}{3} arctan (3 e^{- 2 x})

Simplify

- \frac{1}{2} \cdot \frac{1}{3} arctan (3 e^{- 2 x}) : - \frac{1}{6} arctan (3 e^{- 2 x})

= - \frac{1}{6} arctan (3 e^{- 2 x})

Add a constant to the solution

= - \frac{1}{6} arctan (3 e^{- 2 x}) + C

\frac{d}{d x} (ln (sinh (4 x)))

y^{^{'}} + x^{- 1} y = \frac{2}{3} x^{4} y^{4}

y^{^{''}} + 2 y^{^{'}} + y = 12 e^{- t}

\frac{\partial}{\partial x} (x e^{x} + e^{x})

\frac{\partial}{\partial y} (x - 2 y)

What is the integral of 1/(9e^{-2x)+e^{2x}} ?
The integral of 1/(9e^{-2x)+e^{2x}} is -1/6 arctan(3e^{-2x})+C