What is the integral of 1/(49e^{-7x)+e^{7x}} ?

The integral of 1/(49e^{-7x)+e^{7x}} is -1/49 arctan(7e^{-7x})+C

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integral of 1/(49e^{-7x)+e^{7x}}

\int \frac{1}{49 e ^{- 7 x} + e ^{7 x}} d x

- \frac{1}{49} arctan (7 e^{- 7 x}) + C

Solution steps

\int \frac{1}{49 e ^{- 7 x} + e ^{7 x}} d x

Apply u-substitution

: \int - \frac{1}{7 ( 49 u ^{2} + 1 )} d u

= \int - \frac{1}{7 ( 49 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}{7} \cdot \int \frac{1}{49 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{7} \cdot \int \frac{1}{7 ( 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}{7} \cdot \frac{1}{7} \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}{7} \cdot \frac{1}{7} arctan (v)

Substitute back

= - \frac{1}{7} \cdot \frac{1}{7} arctan (7 e^{- 7 x})

Simplify

- \frac{1}{7} \cdot \frac{1}{7} arctan (7 e^{- 7 x}) : - \frac{1}{49} arctan (7 e^{- 7 x})

= - \frac{1}{49} arctan (7 e^{- 7 x})

Add a constant to the solution

= - \frac{1}{49} arctan (7 e^{- 7 x}) + C

\frac{d y}{d x} = e^{x^{2}}

(ln (x))^{4}

2 x y \frac{d y}{d x} + y^{2} = 2 x^{2}

ln (8 x)

n = 1 \sum \infty \frac{n}{2 n ^{2} + 1}

What is the integral of 1/(49e^{-7x)+e^{7x}} ?
The integral of 1/(49e^{-7x)+e^{7x}} is -1/49 arctan(7e^{-7x})+C