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

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

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integral of (e^{-3x})

\int (e^{- 3 x}) d x

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

Solution steps

\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

\frac{d y}{y} = \frac{4}{x} d x

f (x) = sec (2 x)

x \to 3 lim (\frac{x ^{3} - 8}{x - 2})

\int \frac{x ^{3}}{1 - x ^{2}} d x

\frac{e ^{- s}}{s ^{2}}