What is the integral of 1/(sqrt(9e^{-2x)-1)} ?

The integral of 1/(sqrt(9e^{-2x)-1)} is -arctan(sqrt(9e^{-2x)-1})+C

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integral of 1/(sqrt(9e^{-2x)-1)}

\int \frac{1}{9 e ^{- 2 x} - 1} d x

- arctan (9 e^{- 2 x} - 1) + C

Solution steps

\int \frac{1}{9 e ^{- 2 x} - 1} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (u)

Substitute back

u = 9 e^{- 2 x} - 1

= - arctan (9 e^{- 2 x} - 1)

Add a constant to the solution

= - arctan (9 e^{- 2 x} - 1) + C

What is the integral of 1/(sqrt(9e^{-2x)-1)} ?
The integral of 1/(sqrt(9e^{-2x)-1)} is -arctan(sqrt(9e^{-2x)-1})+C