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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

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 x

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

Add a constant to the solution

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

y d x - x d y + x y d y = 0

n = 0 \sum \infty \frac{2}{n ( n + 2 )}

\int \frac{ln ( x ^{49} )}{x} d x

\frac{\partial}{\partial x} (e^{(x y)^{\frac{1}{2}}})

\int y ln (x) d x

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