Find the Integral xe^(-x)

$x e^{- x}$

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

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

Simplify.

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Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

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

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

Simplify.

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Multiply $- 1$ by $- 1$ .

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

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

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

Let $u = - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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

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Differentiate $- x$ .

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

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

$- 1 \cdot 1$

Multiply $- 1$ by $1$ .

$- 1$

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

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

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

$- x e^{- x} - \int e^{u} d u$

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

$- x e^{- x} - (e^{u} + C)$

Simplify.

$- x e^{- x} - e^{u} + C$

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

$- x e^{- x} - e^{- x} + C$

Find the Integral xe^(-x)

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