Find the Integral xe^(-3x)

$x e^{- 3 x}$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = e^{- 3 x}$ .

$x (- \frac{1}{3} e^{- 3 x}) - \int - \frac{1}{3} e^{- 3 x} d x$

Simplify.

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Combine $e^{- 3 x}$ and $\frac{1}{3}$ .

$x (- \frac{e^{- 3 x}}{3}) - \int - \frac{1}{3} e^{- 3 x} d x$

Combine $x$ and $\frac{e^{- 3 x}}{3}$ .

$- \frac{x e^{- 3 x}}{3} - \int - \frac{1}{3} e^{- 3 x} d x$

Combine $e^{- 3 x}$ and $\frac{1}{3}$ .

$- \frac{x e^{- 3 x}}{3} - \int - \frac{e^{- 3 x}}{3} d x$

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \frac{x e^{- 3 x}}{3} - - \int \frac{e^{- 3 x}}{3} d x$

Simplify.

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Multiply $- 1$ by $- 1$ .

$- \frac{x e^{- 3 x}}{3} + 1 \int \frac{e^{- 3 x}}{3} d x$

Multiply $\int \frac{e^{- 3 x}}{3} d x$ by $1$ .

$- \frac{x e^{- 3 x}}{3} + \int \frac{e^{- 3 x}}{3} d x$

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$- \frac{x e^{- 3 x}}{3} + \frac{1}{3} \int e^{- 3 x} d x$

Let $u = - 3 x$ . Then $d u = - 3 d x$ , so $- \frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = - 3 x$ . Find $\frac{d u}{d x}$ .

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Rewrite.

$\frac{- 3}{1}$

Divide $- 3$ by $1$ .

$- 3$

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

$- \frac{x e^{- 3 x}}{3} + \frac{1}{3} \int e^{u} \frac{1}{- 3} d u$

Simplify.

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Move the negative in front of the fraction.

$- \frac{x e^{- 3 x}}{3} + \frac{1}{3} \int e^{u} (- \frac{1}{3}) d u$

Combine $e^{u}$ and $\frac{1}{3}$ .

$- \frac{x e^{- 3 x}}{3} + \frac{1}{3} \int - \frac{e^{u}}{3} d u$

Since $- 1$ is constant with respect to $u$ , move $- 1$ out of the integral.

$- \frac{x e^{- 3 x}}{3} + \frac{1}{3} (- \int \frac{e^{u}}{3} d u)$

Simplify.

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Combine $\frac{1}{3}$ and $- 1$ .

$- \frac{x e^{- 3 x}}{3} + \frac{- 1}{3} \int \frac{e^{u}}{3} d u$

Move the negative in front of the fraction.

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

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$- \frac{x e^{- 3 x}}{3} - \frac{1}{3} (\frac{1}{3} \int e^{u} d u)$

Simplify.

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Multiply $\frac{1}{3}$ and $\frac{1}{3}$ .

$- \frac{x e^{- 3 x}}{3} - \frac{1}{3 \cdot 3} \int e^{u} d u$

Multiply $3$ by $3$ .

$- \frac{x e^{- 3 x}}{3} - \frac{1}{9} \int e^{u} d u$

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

$- \frac{x e^{- 3 x}}{3} - \frac{1}{9} (e^{u} + C)$

Simplify.

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Rewrite $- \frac{x e^{- 3 x}}{3} - \frac{1}{9} (e^{u} + C)$ as $- \frac{1}{3} x e^{- 3 x} - \frac{1}{9} e^{u} + C$ .

$- \frac{1}{3} x e^{- 3 x} - \frac{1}{9} e^{u} + C$

Simplify.

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Combine $x$ and $\frac{1}{3}$ .

$- \frac{x}{3} e^{- 3 x} - \frac{1}{9} e^{u} + C$

Combine $e^{- 3 x}$ and $\frac{x}{3}$ .

$- \frac{e^{- 3 x} x}{3} - \frac{1}{9} e^{u} + C$

Replace all occurrences of $u$ with $- 3 x$ .

$- \frac{e^{- 3 x} x}{3} - \frac{1}{9} e^{- 3 x} + C$

Combine $e^{- 3 x}$ and $\frac{1}{9}$ .

$- \frac{e^{- 3 x} x}{3} - \frac{e^{- 3 x}}{9} + C$

Reorder terms.

$- \frac{1}{3} e^{- 3 x} x - \frac{1}{9} e^{- 3 x} + C$

Find the Integral xe^(-3x)

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