What is the integral of 1/(25e^{-4x)+e^{4x}} ?

The integral of 1/(25e^{-4x)+e^{4x}} is -1/20 arctan(5e^{-4x})+C

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integral of 1/(25e^{-4x)+e^{4x}}

\int \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

- \frac{1}{20} arctan (5 e^{- 4 x}) + C

Solution steps

\int \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

Apply u-substitution

: \int - \frac{1}{4 ( 25 u ^{2} + 1 )} d u

= \int - \frac{1}{4 ( 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}{4} \cdot \int \frac{1}{25 u ^{2} + 1} d u

Apply Integral Substitution

= - \frac{1}{4} \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}{4} \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}{4} \cdot \frac{1}{5} arctan (v)

Substitute back

= - \frac{1}{4} \cdot \frac{1}{5} arctan (5 e^{- 4 x})

Simplify

- \frac{1}{4} \cdot \frac{1}{5} arctan (5 e^{- 4 x}) : - \frac{1}{20} arctan (5 e^{- 4 x})

= - \frac{1}{20} arctan (5 e^{- 4 x})

Add a constant to the solution

= - \frac{1}{20} arctan (5 e^{- 4 x}) + C

f (x) = (5 x^{6} + 2 x^{3})^{4}

\int \frac{1}{- x - 1} d x

\frac{\partial}{\partial x} (2 (x - y) e^{2 y + 5 x^{2}})

f^{^{'}} (t) - f (t) = - 7 t, f (2) = - 4

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

What is the integral of 1/(25e^{-4x)+e^{4x}} ?
The integral of 1/(25e^{-4x)+e^{4x}} is -1/20 arctan(5e^{-4x})+C