What is the integral of 1/(sqrt(4e^{-2x)-25)} ?

The integral of 1/(sqrt(4e^{-2x)-25)} is -1/5 arctan(1/5 sqrt(4e^{-2x)-25})+C

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integral of 1/(sqrt(4e^{-2x)-25)}

\int \frac{1}{4 e ^{- 2 x} - 25} d x

- \frac{1}{5} arctan (\frac{1}{5} 4 e^{- 2 x} - 25) + C

Solution steps

\int \frac{1}{4 e ^{- 2 x} - 25} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int \frac{2}{v ^{2} + 25} d v

= - \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 25} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + 25} d v

Apply Integral Substitution

= - \frac{1}{2} \cdot 2 \cdot \int \frac{1}{5 ( w ^{2} + 1 )} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot 2 \cdot \frac{1}{5} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= - \frac{1}{2} \cdot 2 \cdot \frac{1}{5} arctan (w)

Substitute back

= - \frac{1}{2} \cdot 2 \cdot \frac{1}{5} arctan (\frac{4 e ^{- 2 x} - 25}{5})

Simplify

- \frac{1}{2} \cdot 2 \cdot \frac{1}{5} arctan (\frac{4 e ^{- 2 x} - 25}{5}) : - \frac{1}{5} arctan (\frac{1}{5} 4 e^{- 2 x} - 25)

= - \frac{1}{5} arctan (\frac{1}{5} 4 e^{- 2 x} - 25)

Add a constant to the solution

= - \frac{1}{5} arctan (\frac{1}{5} 4 e^{- 2 x} - 25) + C

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What is the integral of 1/(sqrt(4e^{-2x)-25)} ?
The integral of 1/(sqrt(4e^{-2x)-25)} is -1/5 arctan(1/5 sqrt(4e^{-2x)-25})+C