Calculus Examples

${(e^{- 2 x} + 1)}^{3}$

Step 1

Simplify.

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Step 1.1

Use the Binomial Theorem.

$\int {(e^{- 2 x})}^{3} + 3 {(e^{- 2 x})}^{2} \cdot 1 + 3 e^{- 2 x} \cdot 1^{2} + 1^{3} d x$

Step 1.2

Simplify each term.

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Step 1.2.1

Multiply the exponents in ${(e^{- 2 x})}^{3}$ .

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Step 1.2.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int e^{- 2 x \cdot 3} + 3 {(e^{- 2 x})}^{2} \cdot 1 + 3 e^{- 2 x} \cdot 1^{2} + 1^{3} d x$

Step 1.2.1.2

Multiply $3$ by $- 2$ .

$\int e^{- 6 x} + 3 {(e^{- 2 x})}^{2} \cdot 1 + 3 e^{- 2 x} \cdot 1^{2} + 1^{3} d x$

Step 1.2.2

Multiply the exponents in ${(e^{- 2 x})}^{2}$ .

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Step 1.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int e^{- 6 x} + 3 e^{- 2 x \cdot 2} \cdot 1 + 3 e^{- 2 x} \cdot 1^{2} + 1^{3} d x$

Step 1.2.2.2

Multiply $2$ by $- 2$ .

$\int e^{- 6 x} + 3 e^{- 4 x} \cdot 1 + 3 e^{- 2 x} \cdot 1^{2} + 1^{3} d x$

Step 1.2.3

Multiply $3$ by $1$ .

$\int e^{- 6 x} + 3 e^{- 4 x} + 3 e^{- 2 x} \cdot 1^{2} + 1^{3} d x$

Step 1.2.4

One to any power is one.

$\int e^{- 6 x} + 3 e^{- 4 x} + 3 e^{- 2 x} \cdot 1 + 1^{3} d x$

Step 1.2.5

Multiply $3$ by $1$ .

$\int e^{- 6 x} + 3 e^{- 4 x} + 3 e^{- 2 x} + 1^{3} d x$

Step 1.2.6

One to any power is one.

$\int e^{- 6 x} + 3 e^{- 4 x} + 3 e^{- 2 x} + 1 d x$

Step 2

Split the single integral into multiple integrals.

$\int e^{- 6 x} d x + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 3

Let $u_{1} = - 6 x$ . Then $d u_{1} = - 6 d x$ , so $- \frac{1}{6} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 3.1

Let $u_{1} = - 6 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $- 6 x$ .

$\frac{d}{d x} [- 6 x]$

Step 3.1.2

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x$ with respect to $x$ is $- 6 \frac{d}{d x} [x]$ .

$- 6 \frac{d}{d x} [x]$

Step 3.1.3

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

$- 6 \cdot 1$

Step 3.1.4

Multiply $- 6$ by $1$ .

$- 6$

Step 3.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int e^{u_{1}} \frac{1}{- 6} d u_{1} + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$\int e^{u_{1}} (- \frac{1}{6}) d u_{1} + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 4.2

Combine $e^{u_{1}}$ and $\frac{1}{6}$ .

$\int - \frac{e^{u_{1}}}{6} d u_{1} + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 5

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

$- \int \frac{e^{u_{1}}}{6} d u_{1} + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 6

Since $\frac{1}{6}$ is constant with respect to $u_{1}$ , move $\frac{1}{6}$ out of the integral.

$- (\frac{1}{6} \int e^{u_{1}} d u_{1}) + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 7

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$- \frac{1}{6} (e^{u_{1}} + C) + \int 3 e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 8

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$- \frac{1}{6} (e^{u_{1}} + C) + 3 \int e^{- 4 x} d x + \int 3 e^{- 2 x} d x + \int d x$

Step 9

Let $u_{2} = - 4 x$ . Then $d u_{2} = - 4 d x$ , so $- \frac{1}{4} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1

Let $u_{2} = - 4 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 9.1.1

Differentiate $- 4 x$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

Multiply $- 4$ by $1$ .

$- 4$

Step 9.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$- \frac{1}{6} (e^{u_{1}} + C) + 3 \int e^{u_{2}} \frac{1}{- 4} d u_{2} + \int 3 e^{- 2 x} d x + \int d x$

Step 10

Simplify.

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Step 10.1

Move the negative in front of the fraction.

$- \frac{1}{6} (e^{u_{1}} + C) + 3 \int e^{u_{2}} (- \frac{1}{4}) d u_{2} + \int 3 e^{- 2 x} d x + \int d x$

Step 10.2

Combine $e^{u_{2}}$ and $\frac{1}{4}$ .

$- \frac{1}{6} (e^{u_{1}} + C) + 3 \int - \frac{e^{u_{2}}}{4} d u_{2} + \int 3 e^{- 2 x} d x + \int d x$

Step 11

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$- \frac{1}{6} (e^{u_{1}} + C) + 3 (- \int \frac{e^{u_{2}}}{4} d u_{2}) + \int 3 e^{- 2 x} d x + \int d x$

Step 12

Multiply $- 1$ by $3$ .

$- \frac{1}{6} (e^{u_{1}} + C) - 3 \int \frac{e^{u_{2}}}{4} d u_{2} + \int 3 e^{- 2 x} d x + \int d x$

Step 13

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

$- \frac{1}{6} (e^{u_{1}} + C) - 3 (\frac{1}{4} \int e^{u_{2}} d u_{2}) + \int 3 e^{- 2 x} d x + \int d x$

Step 14

Simplify.

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Step 14.1

Combine $\frac{1}{4}$ and $- 3$ .

$- \frac{1}{6} (e^{u_{1}} + C) + \frac{- 3}{4} \int e^{u_{2}} d u_{2} + \int 3 e^{- 2 x} d x + \int d x$

Step 14.2

Move the negative in front of the fraction.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} \int e^{u_{2}} d u_{2} + \int 3 e^{- 2 x} d x + \int d x$

Step 15

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + \int 3 e^{- 2 x} d x + \int d x$

Step 16

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + 3 \int e^{- 2 x} d x + \int d x$

Step 17

Let $u_{3} = - 2 x$ . Then $d u_{3} = - 2 d x$ , so $- \frac{1}{2} d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 17.1

Let $u_{3} = - 2 x$ . Find $\frac{d u_{3}}{d x}$ .

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Step 17.1.1

Differentiate $- 2 x$ .

$\frac{d}{d x} [- 2 x]$

Step 17.1.2

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$- 2 \frac{d}{d x} [x]$

Step 17.1.3

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

$- 2 \cdot 1$

Step 17.1.4

Multiply $- 2$ by $1$ .

$- 2$

Step 17.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + 3 \int e^{u_{3}} \frac{1}{- 2} d u_{3} + \int d x$

Step 18

Simplify.

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Step 18.1

Move the negative in front of the fraction.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + 3 \int e^{u_{3}} (- \frac{1}{2}) d u_{3} + \int d x$

Step 18.2

Combine $e^{u_{3}}$ and $\frac{1}{2}$ .

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + 3 \int - \frac{e^{u_{3}}}{2} d u_{3} + \int d x$

Step 19

Since $- 1$ is constant with respect to $u_{3}$ , move $- 1$ out of the integral.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + 3 (- \int \frac{e^{u_{3}}}{2} d u_{3}) + \int d x$

Step 20

Multiply $- 1$ by $3$ .

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) - 3 \int \frac{e^{u_{3}}}{2} d u_{3} + \int d x$

Step 21

Since $\frac{1}{2}$ is constant with respect to $u_{3}$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) - 3 (\frac{1}{2} \int e^{u_{3}} d u_{3}) + \int d x$

Step 22

Simplify.

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Step 22.1

Combine $\frac{1}{2}$ and $- 3$ .

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) + \frac{- 3}{2} \int e^{u_{3}} d u_{3} + \int d x$

Step 22.2

Move the negative in front of the fraction.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) - \frac{3}{2} \int e^{u_{3}} d u_{3} + \int d x$

Step 23

The integral of $e^{u_{3}}$ with respect to $u_{3}$ is $e^{u_{3}}$ .

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) - \frac{3}{2} (e^{u_{3}} + C) + \int d x$

Step 24

Apply the constant rule.

$- \frac{1}{6} (e^{u_{1}} + C) - \frac{3}{4} (e^{u_{2}} + C) - \frac{3}{2} (e^{u_{3}} + C) + x + C$

Step 25

Simplify.

$- \frac{1}{6} e^{u_{1}} - \frac{3}{4} e^{u_{2}} - \frac{3}{2} e^{u_{3}} + x + C$

Step 26

Substitute back in for each integration substitution variable.

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Step 26.1

Replace all occurrences of $u_{1}$ with $- 6 x$ .

$- \frac{1}{6} e^{- 6 x} - \frac{3}{4} e^{u_{2}} - \frac{3}{2} e^{u_{3}} + x + C$

Step 26.2

Replace all occurrences of $u_{2}$ with $- 4 x$ .

$- \frac{1}{6} e^{- 6 x} - \frac{3}{4} e^{- 4 x} - \frac{3}{2} e^{u_{3}} + x + C$

Step 26.3

Replace all occurrences of $u_{3}$ with $- 2 x$ .

$- \frac{1}{6} e^{- 6 x} - \frac{3}{4} e^{- 4 x} - \frac{3}{2} e^{- 2 x} + x + C$