What is the integral of (1+e^{-2y})^{1/4}e^{-2y} ?

The integral of (1+e^{-2y})^{1/4}e^{-2y} is -2/5 (e^{-2y}+1)^{5/4}+C

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integral of (1+e^{-2y})^{1/4}e^{-2y}

\int (1 + e^{- 2 y})^{\frac{1}{4}} e^{- 2 y} d y

- \frac{2}{5} (e^{- 2 y} + 1)^{\frac{5}{4}} + C

Solution steps

\int (1 + e^{- 2 y})^{\frac{1}{4}} e^{- 2 y} d y

Apply u-substitution

: \int - \frac{u ^{\frac{1}{4}}}{2} d u

= \int - \frac{u ^{\frac{1}{4}}}{2} 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 u^{\frac{1}{4}} d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= - \frac{1}{2} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 1 + e^{- 2 y}

= - \frac{1}{2} \cdot \frac{4}{5} (1 + e^{- 2 y})^{\frac{5}{4}}

Simplify

- \frac{1}{2} \cdot \frac{4}{5} (1 + e^{- 2 y})^{\frac{5}{4}} : - \frac{2}{5} (e^{- 2 y} + 1)^{\frac{5}{4}}

= - \frac{2}{5} (e^{- 2 y} + 1)^{\frac{5}{4}}

Add a constant to the solution

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

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