What is the integral of 1/(9e^y+e^{-y)} ?

The integral of 1/(9e^y+e^{-y)} is -1/3 arctan((e^{-y})/3)+C

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integral of 1/(9e^y+e^{-y)}

\int \frac{1}{9 e ^{y} + e ^{- y}} d y

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

Solution steps

\int \frac{1}{9 e ^{y} + e ^{- y}} d y

Apply u-substitution

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

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

Apply Integral Substitution

= - \int \frac{1}{3 ( 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}{3} \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}{3} arctan (v)

Substitute back

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

Add a constant to the solution

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

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What is the integral of 1/(9e^y+e^{-y)} ?
The integral of 1/(9e^y+e^{-y)} is -1/3 arctan((e^{-y})/3)+C