What is the integral of+e^{kt} ?

The integral of+e^{kt} is 1/k e^{kt}+C

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integral of+e^{kt}

\int + e^{k t} d t

\frac{1}{k} e^{k t} + C

Solution steps

\int e^{k t} d t

Apply u-substitution

: \int \frac{e ^{u}}{k} d u

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{k} e^{u}

Substitute back

u = k t

= \frac{1}{k} e^{k t}

Add a constant to the solution

= \frac{1}{k} e^{k t} + C

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