What is the integral of 1/(e^{3y)} ?

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

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integral of 1/(e^{3y)}

\int \frac{1}{e ^{3 y}} d y

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

Solution steps

\int \frac{1}{e ^{3 y}} d y

Apply exponent rule:

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

\frac{1}{e ^{3 y}} = e^{- 3 y}

= \int e^{- 3 y} d y

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= \int - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{3} e^{u}

Substitute back

u = - 3 y

= - \frac{1}{3} e^{- 3 y}

Add a constant to the solution

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

f (r) = 2 π r^{2}

\frac{d}{d x} ((8 x + 7) (x^{2} - 3 x))

\frac{d}{d x} (sin (2 π (1 - x)))

\int \frac{cos ( t )}{7 + sin ( t )} d t

y = x^{2} + 10 x + 25, y = 7 x + 35

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