What is the integral of (e^{-y})/(sqrt(9-16e^{-2y))} ?

The integral of (e^{-y})/(sqrt(9-16e^{-2y))} is -1/4 arcsin(4/3 e^{-y})+C

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integral of (e^{-y})/(sqrt(9-16e^{-2y))}

\int \frac{e ^{- y}}{9 - 16 e ^{- 2 y}} d y

- \frac{1}{4} arcsin (\frac{4}{3} e^{- y}) + C

Solution steps

\int \frac{e ^{- y}}{9 - 16 e ^{- 2 y}} d y

Apply u-substitution

: \int - \frac{1}{9 - 16 u ^{2}} d u

= \int - \frac{1}{9 - 16 u ^{2}} d u

Take the constant out:

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

= - \int \frac{1}{9 - 16 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= - \int \frac{1}{4} d v

Take the constant out:

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

= - \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= - \frac{1}{4} \cdot 1 \cdot v

Substitute back

= - \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} e^{- y})

Simplify

= - \frac{1}{4} arcsin (\frac{4}{3} e^{- y})

Add a constant to the solution

= - \frac{1}{4} arcsin (\frac{4}{3} e^{- y}) + C

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What is the integral of (e^{-y})/(sqrt(9-16e^{-2y))} ?
The integral of (e^{-y})/(sqrt(9-16e^{-2y))} is -1/4 arcsin(4/3 e^{-y})+C