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

The integral of (e^{-2x})/(e^x) is -1/3 e^{-3x}+C

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

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

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

Solution steps

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

Simplify

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

= \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

\frac{d}{d x} (4 x^{\frac{5}{2}})

x \to - 2 lim (\frac{2 + ∣ x ∣}{2 + x})

\frac{d}{d x} (\frac{arcsin ( x )}{sin ( x )})

x \to 0 lim (\frac{x ^{4}}{x})

x \to \frac{3}{7} lim (\frac{∣ 3 - 7 x ∣}{21 x ^{2} - 9 x})

What is the integral of (e^{-2x})/(e^x) ?
The integral of (e^{-2x})/(e^x) is -1/3 e^{-3x}+C