What is the integral of 2/(3e^{-3x)} ?

The integral of 2/(3e^{-3x)} is 2/9 e^{3x}+C

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integral of 2/(3e^{-3x)}

integral

\frac{2}{3 e ^{- 3 x}}

\frac{2}{9} e^{3 x} + C

Solution steps

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

Take the constant out:

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

= \frac{2}{3} \cdot \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}

= \frac{2}{3} \cdot \int e^{3 x} d x

Apply u-substitution

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

= \frac{2}{3} \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{2}{3} \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x

= \frac{2}{3} \cdot \frac{1}{3} e^{3 x}

Simplify

\frac{2}{3} \cdot \frac{1}{3} e^{3 x} : \frac{2}{9} e^{3 x}

= \frac{2}{9} e^{3 x}

Add a constant to the solution

= \frac{2}{9} e^{3 x} + C

\frac{3 x ^{3}}{2}

\frac{1}{( 1 + x ^{\frac{1}{2}} )}

\frac{1}{x ^{4} - 1}

\frac{t ^{3}}{2}

\frac{( 1 + x ) ^{3}}{x ^{2}}

What is the integral of 2/(3e^{-3x)} ?
The integral of 2/(3e^{-3x)} is 2/9 e^{3x}+C