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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int - \frac{1}{3 ( e ^{- u} + e ^{u} )} 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}{e ^{- u} + e ^{u}} d u

Apply u-substitution

: \int \frac{1}{1 + v ^{2}} d v

= - \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

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

Substitute back

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

Add a constant to the solution

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

\int (e^{- x} - e^{5}) (e^{x} + e^{- 3}) d x

\int \frac{1}{3 ( 2 x - 1 ) ^{2}} d x

\int_{0}^{2} e^{x} - e^{- x} d x

g (x) = \frac{4 x ^{2} - x + 2}{3 - 4 x}

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