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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{v ^{2} + 1} d v

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (v)

Substitute back

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

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

Add a constant to the solution

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

2 x^{- 5}

\int - 2 cos (x) d x

\int_{1}^{6} x - 1 d x

y = (13 x^{2} - 26 x + 26) e^{- 8 x}

\int 36 - x^{2} d x

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