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

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

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

\int \frac{e ^{- t} \cdot 2 e ^{- t}}{3 e ^{t}} d t

- \frac{2}{9} e^{- 3 t} + C

Solution steps

\int \frac{e ^{- t} \cdot 2 e ^{- t}}{3 e ^{t}} d t

Take the constant out:

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

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

Simplify

\frac{e ^{- t} e ^{- t}}{e ^{t}} : \frac{1}{e ^{3 t}}

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

Apply exponent rule:

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

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

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

Apply u-substitution

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

= \frac{2}{3} \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

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

Use the common integral:

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

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

Substitute back

u = - 3 t

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

Simplify

\frac{2}{3} (- \frac{1}{3} e^{- 3 t}) : - \frac{2}{9} e^{- 3 t}

= - \frac{2}{9} e^{- 3 t}

Add a constant to the solution

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

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