What is the integral of 1/(64e^{-4x)+e^{4x}} ?

The integral of 1/(64e^{-4x)+e^{4x}} is -1/32 arctan(8e^{-4x})+C

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integral of 1/(64e^{-4x)+e^{4x}}

\int \frac{1}{64 e ^{- 4 x} + e ^{4 x}} d x

- \frac{1}{32} arctan (8 e^{- 4 x}) + C

Solution steps

\int \frac{1}{64 e ^{- 4 x} + e ^{4 x}} d x

Apply u-substitution

: \int - \frac{1}{4 ( 64 u ^{2} + 1 )} d u

= \int - \frac{1}{4 ( 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}{4} \cdot \int \frac{1}{64 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{4} \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}{4} \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}{4} \cdot \frac{1}{8} arctan (v)

Substitute back

= - \frac{1}{4} \cdot \frac{1}{8} arctan (8 e^{- 4 x})

Simplify

- \frac{1}{4} \cdot \frac{1}{8} arctan (8 e^{- 4 x}) : - \frac{1}{32} arctan (8 e^{- 4 x})

= - \frac{1}{32} arctan (8 e^{- 4 x})

Add a constant to the solution

= - \frac{1}{32} arctan (8 e^{- 4 x}) + C

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What is the integral of 1/(64e^{-4x)+e^{4x}} ?
The integral of 1/(64e^{-4x)+e^{4x}} is -1/32 arctan(8e^{-4x})+C