What is the integral of 1/(16e^{-3x)+e^{3x}} ?

The integral of 1/(16e^{-3x)+e^{3x}} is -1/12 arctan(4e^{-3x})+C

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integral of 1/(16e^{-3x)+e^{3x}}

\int \frac{1}{16 e ^{- 3 x} + e ^{3 x}} d x

- \frac{1}{12} arctan (4 e^{- 3 x}) + C

Solution steps

\int \frac{1}{16 e ^{- 3 x} + e ^{3 x}} d x

Apply u-substitution

: \int - \frac{1}{3 ( 16 u ^{2} + 1 )} d u

= \int - \frac{1}{3 ( 16 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}{3} \cdot \int \frac{1}{16 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{3} \cdot \int \frac{1}{4 ( 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}{3} \cdot \frac{1}{4} \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}{3} \cdot \frac{1}{4} arctan (v)

Substitute back

= - \frac{1}{3} \cdot \frac{1}{4} arctan (4 e^{- 3 x})

Simplify

- \frac{1}{3} \cdot \frac{1}{4} arctan (4 e^{- 3 x}) : - \frac{1}{12} arctan (4 e^{- 3 x})

= - \frac{1}{12} arctan (4 e^{- 3 x})

Add a constant to the solution

= - \frac{1}{12} arctan (4 e^{- 3 x}) + C

y = x e^{x}

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 4)^{7})

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

\frac{d}{d x} (2 x^{2} - 5 x + 3)

f (x) = 2 tan^{2} (6 x)

What is the integral of 1/(16e^{-3x)+e^{3x}} ?
The integral of 1/(16e^{-3x)+e^{3x}} is -1/12 arctan(4e^{-3x})+C