What is the integral of (e^{kx})/k ?

The integral of (e^{kx})/k is e^{kx} 1/(k^2)+C

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integral of (e^{kx})/k

\int \frac{e ^{k x}}{k} d x

e^{k x} \frac{1}{k ^{2}} + C

Solution steps

\int \frac{e ^{k x}}{k} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = k x

= \frac{1}{k} \cdot \frac{1}{k} e^{k x}

Simplify

\frac{1}{k} \cdot \frac{1}{k} e^{k x} : e^{k x} \frac{1}{k ^{2}}

= e^{k x} \frac{1}{k ^{2}}

Add a constant to the solution

= e^{k x} \frac{1}{k ^{2}} + C

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What is the integral of (e^{kx})/k ?
The integral of (e^{kx})/k is e^{kx} 1/(k^2)+C