What is the integral of 1/(36e^{-3x)+e^{3x}} ?

The integral of 1/(36e^{-3x)+e^{3x}} is -1/18 arctan(6e^{-3x})+C

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integral of 1/(36e^{-3x)+e^{3x}}

\int \frac{1}{36 e ^{- 3 x} + e ^{3 x}} d x

- \frac{1}{18} arctan (6 e^{- 3 x}) + C

Solution steps

\int \frac{1}{36 e ^{- 3 x} + e ^{3 x}} d x

Apply u-substitution

: \int - \frac{1}{3 ( 36 u ^{2} + 1 )} d u

= \int - \frac{1}{3 ( 36 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}{36 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{3} \cdot \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

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

Simplify

- \frac{1}{3} \cdot \frac{1}{6} arctan (6 e^{- 3 x}) : - \frac{1}{18} arctan (6 e^{- 3 x})

= - \frac{1}{18} arctan (6 e^{- 3 x})

Add a constant to the solution

= - \frac{1}{18} arctan (6 e^{- 3 x}) + C

\frac{d}{d x} (tan (x) - sin (x))

h (x) = - 6 x^{- 2}

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

\int \frac{3 r ^{2}}{r + 7} d r

\frac{s}{s ^{2} - 2 s - 3}

What is the integral of 1/(36e^{-3x)+e^{3x}} ?
The integral of 1/(36e^{-3x)+e^{3x}} is -1/18 arctan(6e^{-3x})+C