Calculus Examples

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

Step 1

Simplify.

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Step 1.1

Rewrite ${(e^{3 x} - 2)}^{2}$ as $(e^{3 x} - 2) (e^{3 x} - 2)$ .

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

Step 1.2

Expand $(e^{3 x} - 2) (e^{3 x} - 2)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

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

Step 1.2.2

Apply the distributive property.

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

Step 1.2.3

Apply the distributive property.

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

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $e^{3 x}$ by $e^{3 x}$ by adding the exponents.

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Step 1.3.1.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.3.1.1.2

Add $3 x$ and $3 x$ .

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

Step 1.3.1.2

Move $- 2$ to the left of $e^{3 x}$ .

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

Step 1.3.1.3

Multiply $- 2$ by $- 2$ .

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

Step 1.3.2

Subtract $2 e^{3 x}$ from $- 2 e^{3 x}$ .

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

Step 2

Split the single integral into multiple integrals.

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

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 $d u_{1}$ .

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

Step 4

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

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

Step 5

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

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

Step 6

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

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

Step 7

Since $- 4$ is constant with respect to $x$ , move $- 4$ out of the integral.

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

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $3 x$ .

$\frac{d}{d x} [3 x]$

Step 8.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x]$

Step 8.1.3

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

$3 \cdot 1$

Step 8.1.4

Multiply $3$ by $1$ .

$3$

Step 8.2

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

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

Step 9

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

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

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

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

Step 11.2

Move the negative in front of the fraction.

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

Step 12

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

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

Step 13

Apply the constant rule.

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

Step 14

Simplify.

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

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

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

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

Step 15.2

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

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