What is the integral of 1/(e^{3t)} ?

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

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integral of 1/(e^{3t)}

\int \frac{1}{e ^{3 t}} d t

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

Solution steps

\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}

= \int e^{- 3 t} d t

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 t

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

Add a constant to the solution

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

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