What is the integral of 1/(25e^{-5x)+e^{5x}} ?

The integral of 1/(25e^{-5x)+e^{5x}} is -1/25 arctan(5e^{-5x})+C

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integral of 1/(25e^{-5x)+e^{5x}}

\int \frac{1}{25 e ^{- 5 x} + e ^{5 x}} d x

- \frac{1}{25} arctan (5 e^{- 5 x}) + C

Solution steps

\int \frac{1}{25 e ^{- 5 x} + e ^{5 x}} d x

Apply u-substitution

: \int - \frac{1}{5 ( 25 u ^{2} + 1 )} d u

= \int - \frac{1}{5 ( 25 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}{25 u ^{2} + 1} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

- \frac{1}{5} \cdot \frac{1}{5} arctan (5 e^{- 5 x}) : - \frac{1}{25} arctan (5 e^{- 5 x})

= - \frac{1}{25} arctan (5 e^{- 5 x})

Add a constant to the solution

= - \frac{1}{25} arctan (5 e^{- 5 x}) + C

x \to \frac{π}{4} lim (\frac{1}{cos ( x )})

f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 4)

\int_{0}^{2} (x^{3} - 1)^{2} d x

y^{^{'}} = y - 2 x - 1

x + 2 y x^{2} + 1 y^{^{'}} = 0

What is the integral of 1/(25e^{-5x)+e^{5x}} ?
The integral of 1/(25e^{-5x)+e^{5x}} is -1/25 arctan(5e^{-5x})+C