Evaluate integral of e^(kx) with respect to x

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

Let $u = k x$ . Then $d u = k d x$ , so $\frac{1}{k} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = k x$ . Find $\frac{d u}{d x}$ .

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Differentiate $k x$ .

$\frac{d}{d x} [k x]$

Since $k$ is constant with respect to $x$ , the derivative of $k x$ with respect to $x$ is $k \frac{d}{d x} [x]$ .

$k \frac{d}{d x} [x]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$k \cdot 1$

Multiply $k$ by $1$ .

$k$

Rewrite the problem using $u$ and $d u$ .

$\int e^{u} \frac{1}{k} d u$

Combine $e^{u}$ and $\frac{1}{k}$ .

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

Since $\frac{1}{k}$ is constant with respect to $u$ , move $\frac{1}{k}$ out of the integral.

$\frac{1}{k} \int e^{u} d u$

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$\frac{1}{k} (e^{u} + C)$

Simplify.

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Simplify.

$\frac{1}{k} e^{u} + C$

Combine $\frac{1}{k}$ and $e^{u}$ .

$\frac{e^{u}}{k} + C$

Replace all occurrences of $u$ with $k x$ .

$\frac{e^{k x}}{k} + C$

Evaluate integral of e^(kx) with respect to x

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