Calculus Examples

$\int e^{8 x} + e^{- 9 x} d x$

Step 1

Split the single integral into multiple integrals.

$\int e^{8 x} d x + \int e^{- 9 x} d x$

Step 2

Let $u_{1} = 8 x$ . Then $d u_{1} = 8 d x$ , so $\frac{1}{8} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.1

Let $u_{1} = 8 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 2.1.1

Differentiate $8 x$ .

$\frac{d}{d x} [8 x]$

Step 2.1.2

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$8 \frac{d}{d x} [x]$

Step 2.1.3

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

$8 \cdot 1$

Step 2.1.4

Multiply $8$ by $1$ .

$8$

Step 2.2

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

$\int e^{u_{1}} \frac{1}{8} d u_{1} + \int e^{- 9 x} d x$

Step 3

Combine $e^{u_{1}}$ and $\frac{1}{8}$ .

$\int \frac{e^{u_{1}}}{8} d u_{1} + \int e^{- 9 x} d x$

Step 4

Since $\frac{1}{8}$ is constant with respect to $u_{1}$ , move $\frac{1}{8}$ out of the integral.

$\frac{1}{8} \int e^{u_{1}} d u_{1} + \int e^{- 9 x} d x$

Step 5

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

$\frac{1}{8} (e^{u_{1}} + C) + \int e^{- 9 x} d x$

Step 6

Let $u_{2} = - 9 x$ . Then $d u_{2} = - 9 d x$ , so $- \frac{1}{9} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 6.1

Let $u_{2} = - 9 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 6.1.1

Differentiate $- 9 x$ .

$\frac{d}{d x} [- 9 x]$

Step 6.1.2

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$- 9 \frac{d}{d x} [x]$

Step 6.1.3

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

$- 9 \cdot 1$

Step 6.1.4

Multiply $- 9$ by $1$ .

$- 9$

Step 6.2

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

$\frac{1}{8} (e^{u_{1}} + C) + \int e^{u_{2}} \frac{1}{- 9} d u_{2}$

Step 7

Simplify.

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Step 7.1

Move the negative in front of the fraction.

$\frac{1}{8} (e^{u_{1}} + C) + \int e^{u_{2}} (- \frac{1}{9}) d u_{2}$

Step 7.2

Combine $e^{u_{2}}$ and $\frac{1}{9}$ .

$\frac{1}{8} (e^{u_{1}} + C) + \int - \frac{e^{u_{2}}}{9} d u_{2}$

Step 8

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

$\frac{1}{8} (e^{u_{1}} + C) - \int \frac{e^{u_{2}}}{9} d u_{2}$

Step 9

Since $\frac{1}{9}$ is constant with respect to $u_{2}$ , move $\frac{1}{9}$ out of the integral.

$\frac{1}{8} (e^{u_{1}} + C) - (\frac{1}{9} \int e^{u_{2}} d u_{2})$

Step 10

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

$\frac{1}{8} (e^{u_{1}} + C) - \frac{1}{9} (e^{u_{2}} + C)$

Step 11

Simplify.

$\frac{1}{8} e^{u_{1}} - \frac{1}{9} e^{u_{2}} + C$

Step 12

Substitute back in for each integration substitution variable.

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Step 12.1

Replace all occurrences of $u_{1}$ with $8 x$ .

$\frac{1}{8} e^{8 x} - \frac{1}{9} e^{u_{2}} + C$

Step 12.2

Replace all occurrences of $u_{2}$ with $- 9 x$ .

$\frac{1}{8} e^{8 x} - \frac{1}{9} e^{- 9 x} + C$