What is the integral of 1/(36e^{-5x)+e^{5x}} ?

The integral of 1/(36e^{-5x)+e^{5x}} is -1/30 arctan(6e^{-5x})+C

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integral of 1/(36e^{-5x)+e^{5x}}

\int \frac{1}{36 e ^{- 5 x} + e ^{5 x}} d x

- \frac{1}{30} arctan (6 e^{- 5 x}) + C

Solution steps

\int \frac{1}{36 e ^{- 5 x} + e ^{5 x}} d x

Apply u-substitution

: \int - \frac{1}{5 ( 36 u ^{2} + 1 )} d u

= \int - \frac{1}{5 ( 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}{5} \cdot \int \frac{1}{36 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{5} \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}{5} \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}{5} \cdot \frac{1}{6} arctan (v)

Substitute back

= - \frac{1}{5} \cdot \frac{1}{6} arctan (6 e^{- 5 x})

Simplify

- \frac{1}{5} \cdot \frac{1}{6} arctan (6 e^{- 5 x}) : - \frac{1}{30} arctan (6 e^{- 5 x})

= - \frac{1}{30} arctan (6 e^{- 5 x})

Add a constant to the solution

= - \frac{1}{30} arctan (6 e^{- 5 x}) + C

(4) (8.8)

\int 9 (9 x + 2) d x

sin (2 x), 0.0001

x \to 3 lim ([\frac{8}{8 + 10}])

\frac{\partial}{\partial y} (tan (\frac{y}{x}))

What is the integral of 1/(36e^{-5x)+e^{5x}} ?
The integral of 1/(36e^{-5x)+e^{5x}} is -1/30 arctan(6e^{-5x})+C