Calculus Examples

$\int (e^{3 x} - 18 e^{- 4 x}) d x$

Step 1

Remove parentheses.

$\int e^{3 x} - 18 e^{- 4 x} d x$

Step 2

Split the single integral into multiple integrals.

$\int e^{3 x} d x + \int - 18 e^{- 4 x} d x$

Step 3

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

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Step 3.1

Let $u_{1} = 3 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $3 x$ .

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

Step 3.1.2

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 3.1.3

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

$3 \cdot 1$

Step 3.1.4

Multiply $3$ by $1$ .

$3$

Step 3.2

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

$\int e^{u_{1}} \frac{1}{3} d u_{1} + \int - 18 e^{- 4 x} d x$

Step 4

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

$\int \frac{e^{u_{1}}}{3} d u_{1} + \int - 18 e^{- 4 x} d x$

Step 5

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

$\frac{1}{3} \int e^{u_{1}} d u_{1} + \int - 18 e^{- 4 x} d x$

Step 6

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

$\frac{1}{3} (e^{u_{1}} + C) + \int - 18 e^{- 4 x} d x$

Step 7

Since $- 18$ is constant with respect to $x$ , move $- 18$ out of the integral.

$\frac{1}{3} (e^{u_{1}} + C) - 18 \int e^{- 4 x} d x$

Step 8

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 8.1

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

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Step 8.1.1

Differentiate $- 4 x$ .

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

Step 8.1.2

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 8.1.3

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

$- 4 \cdot 1$

Step 8.1.4

Multiply $- 4$ by $1$ .

$- 4$

Step 8.2

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

$\frac{1}{3} (e^{u_{1}} + C) - 18 \int e^{u_{2}} \frac{1}{- 4} d u_{2}$

Step 9

Simplify.

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Step 9.1

Move the negative in front of the fraction.

$\frac{1}{3} (e^{u_{1}} + C) - 18 \int e^{u_{2}} (- \frac{1}{4}) d u_{2}$

Step 9.2

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

$\frac{1}{3} (e^{u_{1}} + C) - 18 \int - \frac{e^{u_{2}}}{4} d u_{2}$

Step 10

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

$\frac{1}{3} (e^{u_{1}} + C) - 18 (- \int \frac{e^{u_{2}}}{4} d u_{2})$

Step 11

Multiply $- 1$ by $- 18$ .

$\frac{1}{3} (e^{u_{1}} + C) + 18 \int \frac{e^{u_{2}}}{4} d u_{2}$

Step 12

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

$\frac{1}{3} (e^{u_{1}} + C) + 18 (\frac{1}{4} \int e^{u_{2}} d u_{2})$

Step 13

Simplify.

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Step 13.1

Combine $\frac{1}{4}$ and $18$ .

$\frac{1}{3} (e^{u_{1}} + C) + \frac{18}{4} \int e^{u_{2}} d u_{2}$

Step 13.2

Cancel the common factor of $18$ and $4$ .

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Step 13.2.1

Factor $2$ out of $18$ .

$\frac{1}{3} (e^{u_{1}} + C) + \frac{2 (9)}{4} \int e^{u_{2}} d u_{2}$

Step 13.2.2

Cancel the common factors.

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Step 13.2.2.1

Factor $2$ out of $4$ .

$\frac{1}{3} (e^{u_{1}} + C) + \frac{2 \cdot 9}{2 \cdot 2} \int e^{u_{2}} d u_{2}$

Step 13.2.2.2

Cancel the common factor.

$\frac{1}{3} (e^{u_{1}} + C) + \frac{2 \cdot 9}{2 \cdot 2} \int e^{u_{2}} d u_{2}$

Step 13.2.2.3

Rewrite the expression.

$\frac{1}{3} (e^{u_{1}} + C) + \frac{9}{2} \int e^{u_{2}} d u_{2}$

Step 14

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

$\frac{1}{3} (e^{u_{1}} + C) + \frac{9}{2} (e^{u_{2}} + C)$

Step 15

Simplify.

$\frac{1}{3} e^{u_{1}} + \frac{9}{2} e^{u_{2}} + C$

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

Replace all occurrences of $u_{1}$ with $3 x$ .

$\frac{1}{3} e^{3 x} + \frac{9}{2} e^{u_{2}} + C$

Step 16.2

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

$\frac{1}{3} e^{3 x} + \frac{9}{2} e^{- 4 x} + C$