What is the integral of 1/(e^{5y)} ?

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

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integral of 1/(e^{5y)}

\int \frac{1}{e ^{5 y}} d y

- \frac{1}{5} e^{- 5 y} + C

Solution steps

\int \frac{1}{e ^{5 y}} d y

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{5 y}} = e^{- 5 y}

= \int e^{- 5 y} d y

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= \int - \frac{1}{5} e^{u} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{5} e^{u}

Substitute back

u = - 5 y

= - \frac{1}{5} e^{- 5 y}

Add a constant to the solution

= - \frac{1}{5} e^{- 5 y} + C

\frac{4 s + 1}{s ^{3} - 3 s ^{2} + 3 s - 1}

\frac{( 10 s - 3 )}{s ^{2} - 4 s - 29}

\frac{d}{d x} (\frac{ln ( x )}{x})

\frac{\partial}{\partial y} (9 x^{y} + 2 y^{x} + 11)

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