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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \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}

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

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

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

= 2 (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

= 2 (- \frac{1}{3} e^{- 3 x})

Simplify

2 (- \frac{1}{3} e^{- 3 x}) : - \frac{2}{3} e^{- 3 x}

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

Add a constant to the solution

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

\frac{d}{d x} (4^{s i n (2) (x) + c o s (3 x)})

\int \frac{1}{1 - 4 t ^{2}} d t

y^{^{'}} = 0.2 ln (\frac{4000}{y}) y

\int_{0}^{1} 2

\int [x^{2} + \frac{1}{( 3 x ) ^{2}}] d x

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