What is the integral of 1/(64e^{-3x)+e^{3x}} ?

The integral of 1/(64e^{-3x)+e^{3x}} is -1/24 arctan(8e^{-3x})+C

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integral of 1/(64e^{-3x)+e^{3x}}

\int \frac{1}{64 e ^{- 3 x} + e ^{3 x}} d x

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

Solution steps

\int \frac{1}{64 e ^{- 3 x} + e ^{3 x}} d x

Apply u-substitution

: \int - \frac{1}{3 ( 64 u ^{2} + 1 )} d u

= \int - \frac{1}{3 ( 64 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}{64 u ^{2} + 1} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty sin (nπ)

(2 x^{3} - x y^{2} - 2 y + 3) d x - (x^{2} y + 2 x) d y = 0

t^{2} e^{3 t}

\int sin^{2} (4 x) cos^{2} (4 x) d x

\frac{d}{d x} (5^{s i n (π x)})

What is the integral of 1/(64e^{-3x)+e^{3x}} ?
The integral of 1/(64e^{-3x)+e^{3x}} is -1/24 arctan(8e^{-3x})+C