Calculus Examples

$\int 5 x e^{- 4 x} d x$

Step 1

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$5 \int x e^{- 4 x} d x$

Step 2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = e^{- 4 x}$ .

$5 (x (- \frac{1}{4} e^{- 4 x}) - \int - \frac{1}{4} e^{- 4 x} d x)$

Step 3

Simplify.

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Step 3.1

Combine $e^{- 4 x}$ and $\frac{1}{4}$ .

$5 (x (- \frac{e^{- 4 x}}{4}) - \int - \frac{1}{4} e^{- 4 x} d x)$

Step 3.2

Combine $x$ and $\frac{e^{- 4 x}}{4}$ .

$5 (- \frac{x e^{- 4 x}}{4} - \int - \frac{1}{4} e^{- 4 x} d x)$

Step 3.3

Combine $e^{- 4 x}$ and $\frac{1}{4}$ .

$5 (- \frac{x e^{- 4 x}}{4} - \int - \frac{e^{- 4 x}}{4} d x)$

Step 4

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

$5 (- \frac{x e^{- 4 x}}{4} - - \int \frac{e^{- 4 x}}{4} d x)$

Step 5

Simplify.

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Step 5.1

Multiply $- 1$ by $- 1$ .

$5 (- \frac{x e^{- 4 x}}{4} + 1 \int \frac{e^{- 4 x}}{4} d x)$

Step 5.2

Multiply $\int \frac{e^{- 4 x}}{4} d x$ by $1$ .

$5 (- \frac{x e^{- 4 x}}{4} + \int \frac{e^{- 4 x}}{4} d x)$

Step 6

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$5 (- \frac{x e^{- 4 x}}{4} + \frac{1}{4} \int e^{- 4 x} d x)$

Step 7

Let $u = - 4 x$ . Then $d u = - 4 d x$ , so $- \frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 7.1

Let $u = - 4 x$ . Find $\frac{d u}{d x}$ .

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Step 7.1.1

Differentiate $- 4 x$ .

$\frac{d}{d x} [- 4 x]$

Step 7.1.2

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$- 4 \frac{d}{d x} [x]$

Step 7.1.3

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

$- 4 \cdot 1$

Step 7.1.4

Multiply $- 4$ by $1$ .

$- 4$

Step 7.2

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

$5 (- \frac{x e^{- 4 x}}{4} + \frac{1}{4} \int e^{u} \frac{1}{- 4} d u)$

Step 8

Simplify.

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Step 8.1

Move the negative in front of the fraction.

$5 (- \frac{x e^{- 4 x}}{4} + \frac{1}{4} \int e^{u} (- \frac{1}{4}) d u)$

Step 8.2

Combine $e^{u}$ and $\frac{1}{4}$ .

$5 (- \frac{x e^{- 4 x}}{4} + \frac{1}{4} \int - \frac{e^{u}}{4} d u)$

Step 9

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

$5 (- \frac{x e^{- 4 x}}{4} + \frac{1}{4} (- \int \frac{e^{u}}{4} d u))$

Step 10

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$5 (- \frac{x e^{- 4 x}}{4} + \frac{1}{4} (- (\frac{1}{4} \int e^{u} d u)))$

Step 11

Simplify.

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Step 11.1

Multiply $\frac{1}{4}$ by $\frac{1}{4}$ .

$5 (- \frac{x e^{- 4 x}}{4} - \frac{1}{4 \cdot 4} \int e^{u} d u)$

Step 11.2

Multiply $4$ by $4$ .

$5 (- \frac{x e^{- 4 x}}{4} - \frac{1}{16} \int e^{u} d u)$

Step 12

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

$5 (- \frac{x e^{- 4 x}}{4} - \frac{1}{16} (e^{u} + C))$

Step 13

Simplify.

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Step 13.1

Rewrite $5 (- \frac{x e^{- 4 x}}{4} - \frac{1}{16} (e^{u} + C))$ as $5 (- \frac{1}{4} x e^{- 4 x} - \frac{1}{16} e^{u}) + C$ .

$5 (- \frac{1}{4} x e^{- 4 x} - \frac{1}{16} e^{u}) + C$

Step 13.2

Simplify.

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Step 13.2.1

Combine $x$ and $\frac{1}{4}$ .

$5 (- \frac{x}{4} e^{- 4 x} - \frac{1}{16} e^{u}) + C$

Step 13.2.2

Combine $e^{- 4 x}$ and $\frac{x}{4}$ .

$5 (- \frac{e^{- 4 x} x}{4} - \frac{1}{16} e^{u}) + C$

Step 13.2.3

Combine $e^{u}$ and $\frac{1}{16}$ .

$5 (- \frac{e^{- 4 x} x}{4} - \frac{e^{u}}{16}) + C$

Step 14

Replace all occurrences of $u$ with $- 4 x$ .

$5 (- \frac{e^{- 4 x} x}{4} - \frac{e^{- 4 x}}{16}) + C$

Step 15

Reorder terms.

$5 (- \frac{1}{4} e^{- 4 x} x - \frac{1}{16} e^{- 4 x}) + C$