Calculus Examples

$\int_{0}^{5} \sqrt{x^{2} + 25} d x$

Step 1

Let $x = 5 \tan (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = 5 \sec^{2} (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $5 \sec^{2} (t)$ is positive.

$\int_{0}^{\frac{π}{4}} \sqrt{{(5 \tan (t))}^{2} + 25} (5 \sec^{2} (t)) d t$

Step 2

Simplify terms.

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Step 2.1

Simplify $\sqrt{{(5 \tan (t))}^{2} + 25}$ .

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

Apply the product rule to $5 \tan (t)$ .

$\int_{0}^{\frac{π}{4}} \sqrt{5^{2} \tan^{2} (t) + 25} (5 \sec^{2} (t)) d t$

Step 2.1.1.2

Raise $5$ to the power of $2$ .

$\int_{0}^{\frac{π}{4}} \sqrt{25 \tan^{2} (t) + 25} (5 \sec^{2} (t)) d t$

Step 2.1.2

Factor $25$ out of $25 \tan^{2} (t) + 25$ .

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Step 2.1.2.1

Factor $25$ out of $25 \tan^{2} (t)$ .

$\int_{0}^{\frac{π}{4}} \sqrt{25 (\tan^{2} (t)) + 25} (5 \sec^{2} (t)) d t$

Step 2.1.2.2

Factor $25$ out of $25$ .

$\int_{0}^{\frac{π}{4}} \sqrt{25 (\tan^{2} (t)) + 25 (1)} (5 \sec^{2} (t)) d t$

Step 2.1.2.3

Factor $25$ out of $25 (\tan^{2} (t)) + 25 (1)$ .

$\int_{0}^{\frac{π}{4}} \sqrt{25 (\tan^{2} (t) + 1)} (5 \sec^{2} (t)) d t$

Step 2.1.3

Apply pythagorean identity.

$\int_{0}^{\frac{π}{4}} \sqrt{25 \sec^{2} (t)} (5 \sec^{2} (t)) d t$

Step 2.1.4

Rewrite $25 \sec^{2} (t)$ as ${(5 \sec (t))}^{2}$ .

$\int_{0}^{\frac{π}{4}} \sqrt{{(5 \sec (t))}^{2}} (5 \sec^{2} (t)) d t$

Step 2.1.5

Pull terms out from under the radical, assuming positive real numbers.

$\int_{0}^{\frac{π}{4}} 5 \sec (t) (5 \sec^{2} (t)) d t$

Step 2.2

Simplify.

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Step 2.2.1

Multiply $5$ by $5$ .

$\int_{0}^{\frac{π}{4}} 25 \sec (t) \sec^{2} (t) d t$

Step 2.2.2

Multiply $\sec (t)$ by $\sec^{2} (t)$ by adding the exponents.

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Step 2.2.2.1

Move $\sec^{2} (t)$ .

$\int_{0}^{\frac{π}{4}} 25 (\sec^{2} (t) \sec (t)) d t$

Step 2.2.2.2

Multiply $\sec^{2} (t)$ by $\sec (t)$ .

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Step 2.2.2.2.1

Raise $\sec (t)$ to the power of $1$ .

$\int_{0}^{\frac{π}{4}} 25 (\sec^{2} (t) \sec^{1} (t)) d t$

Step 2.2.2.2.2

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

$\int_{0}^{\frac{π}{4}} 25 {\sec (t)}^{2 + 1} d t$

Step 2.2.2.3

Add $2$ and $1$ .

$\int_{0}^{\frac{π}{4}} 25 \sec^{3} (t) d t$

Step 3

Since $25$ is constant with respect to $t$ , move $25$ out of the integral.

$25 \int_{0}^{\frac{π}{4}} \sec^{3} (t) d t$

Step 4

Apply the reduction formula.

$25 (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \int_{0}^{\frac{π}{4}} \sec (t) d t)$

Step 5

The integral of $\sec (t)$ with respect to $t$ is $\ln (| \sec (t) + \tan (t) |)$ .

$25 (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}})$

Step 6

Simplify.

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Step 6.1

Combine $\frac{1}{2}$ and $\ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}$ .

$25 (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}} + \frac{\ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}}{2})$

Step 6.2

To write $\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$25 (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}} \cdot \frac{2}{2} + \frac{\ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}}{2})$

Step 6.3

Combine $\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}}$ and $\frac{2}{2}$ .

$25 (\frac{\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}} \cdot 2}{2} + \frac{\ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}}{2})$

Step 6.4

Combine the numerators over the common denominator.

$25 \frac{\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}} \cdot 2 + \ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}}{2}$

Step 6.5

Move $2$ to the left of $\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}}$ .

$25 \frac{2 \cdot (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}}) + \ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}}{2}$

Step 6.6

Combine $25$ and $\frac{2 (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}}) + \ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}}}{2}$ .

$\frac{25 (2 (\frac{\tan (t) \sec (t)}{2}]_{0}^{\frac{π}{4}}) + \ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}})}{2}$

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $\frac{\tan (t) \sec (t)}{2}$ at $\frac{π}{4}$ and at $0$ .

$\frac{25 (2 ((\frac{\tan (\frac{π}{4}) \sec (\frac{π}{4})}{2}) - \frac{\tan (0) \sec (0)}{2}) + \ln (| \sec (t) + \tan (t) |)]_{0}^{\frac{π}{4}})}{2}$

Step 7.2

Evaluate $\ln (| \sec (t) + \tan (t) |)$ at $\frac{π}{4}$ and at $0$ .

$\frac{25 (2 ((\frac{\tan (\frac{π}{4}) \sec (\frac{π}{4})}{2}) - \frac{\tan (0) \sec (0)}{2}) + (\ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |)) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 7.3

Remove unnecessary parentheses.

$\frac{25 (2 (\frac{\tan (\frac{π}{4}) \sec (\frac{π}{4})}{2} - \frac{\tan (0) \sec (0)}{2}) + \ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8

Simplify.

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Step 8.1

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$\frac{25 (2 (\frac{1 \sec (\frac{π}{4})}{2} - \frac{\tan (0) \sec (0)}{2}) + \ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8.2

The exact value of $\sec (\frac{π}{4})$ is $\frac{2}{\sqrt{2}}$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{\tan (0) \sec (0)}{2}) + \ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8.3

The exact value of $\tan (0)$ is $0$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{0 \sec (0)}{2}) + \ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8.4

The exact value of $\sec (0)$ is $1$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{0 \cdot 1}{2}) + \ln (| \sec (\frac{π}{4}) + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8.5

The exact value of $\sec (\frac{π}{4})$ is $\frac{2}{\sqrt{2}}$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + \tan (\frac{π}{4}) |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8.6

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| \sec (0) + \tan (0) |))}{2}$

Step 8.7

The exact value of $\sec (0)$ is $1$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + \tan (0) |))}{2}$

Step 8.8

The exact value of $\tan (0)$ is $0$ .

$\frac{25 (2 (\frac{1 \frac{2}{\sqrt{2}}}{2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.9

Multiply $\frac{2}{\sqrt{2}}$ by $1$ .

$\frac{25 (2 (\frac{\frac{2}{\sqrt{2}}}{2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.10

Rewrite $\frac{\frac{2}{\sqrt{2}}}{2}$ as a product.

$\frac{25 (2 (\frac{2}{\sqrt{2}} \cdot \frac{1}{2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.11

Multiply $\frac{2}{\sqrt{2}}$ by $\frac{1}{2}$ .

$\frac{25 (2 (\frac{2}{\sqrt{2} \cdot 2} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.12

Move $2$ to the left of $\sqrt{2}$ .

$\frac{25 (2 (\frac{2}{2 \cdot \sqrt{2}} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.13

Cancel the common factor of $2$ .

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Step 8.13.1

Cancel the common factor.

$\frac{25 (2 (\frac{2}{2 \sqrt{2}} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.13.2

Rewrite the expression.

$\frac{25 (2 (\frac{1}{\sqrt{2}} - \frac{0 \cdot 1}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.14

Multiply $0$ by $1$ .

$\frac{25 (2 (\frac{1}{\sqrt{2}} - \frac{0}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.15

Cancel the common factor of $0$ and $2$ .

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Step 8.15.1

Factor $2$ out of $0$ .

$\frac{25 (2 (\frac{1}{\sqrt{2}} - \frac{2 (0)}{2}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.15.2

Cancel the common factors.

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Step 8.15.2.1

Factor $2$ out of $2$ .

$\frac{25 (2 (\frac{1}{\sqrt{2}} - \frac{2 \cdot 0}{2 \cdot 1}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.15.2.2

Cancel the common factor.

$\frac{25 (2 (\frac{1}{\sqrt{2}} - \frac{2 \cdot 0}{2 \cdot 1}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.15.2.3

Rewrite the expression.

$\frac{25 (2 (\frac{1}{\sqrt{2}} - \frac{0}{1}) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.15.2.4

Divide $0$ by $1$ .

$\frac{25 (2 (\frac{1}{\sqrt{2}} - 0) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.16

Multiply $- 1$ by $0$ .

$\frac{25 (2 (\frac{1}{\sqrt{2}} + 0) + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.17

Add $\frac{1}{\sqrt{2}}$ and $0$ .

$\frac{25 (2 \frac{1}{\sqrt{2}} + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.18

Combine $2$ and $\frac{1}{\sqrt{2}}$ .

$\frac{25 (\frac{2}{\sqrt{2}} + \ln (| \frac{2}{\sqrt{2}} + 1 |) - \ln (| 1 + 0 |))}{2}$

Step 8.19

To write $\ln (| \frac{2}{\sqrt{2}} + 1 |)$ as a fraction with a common denominator, multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{25 (\frac{2}{\sqrt{2}} + \frac{\ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2}}{\sqrt{2}} - \ln (| 1 + 0 |))}{2}$

Step 8.20

Combine the numerators over the common denominator.

$\frac{25 (\frac{2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2}}{\sqrt{2}} - \ln (| 1 + 0 |))}{2}$

Step 8.21

Add $1$ and $0$ .

$\frac{25 (\frac{2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2}}{\sqrt{2}} - \ln (| 1 |))}{2}$

Step 8.22

To write $- \ln (| 1 |)$ as a fraction with a common denominator, multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{25 (\frac{2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2}}{\sqrt{2}} - \ln (| 1 |) \cdot \frac{\sqrt{2}}{\sqrt{2}})}{2}$

Step 8.23

Combine $- \ln (| 1 |)$ and $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{25 (\frac{2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2}}{\sqrt{2}} + \frac{- \ln (| 1 |) \sqrt{2}}{\sqrt{2}})}{2}$

Step 8.24

Combine the numerators over the common denominator.

$\frac{25 \frac{2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2}}{\sqrt{2}}}{2}$

Step 8.25

Combine $25$ and $\frac{2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2}}{\sqrt{2}}$ .

$\frac{\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{\sqrt{2}}}{2}$

Step 8.26

Rewrite $\frac{\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{\sqrt{2}}}{2}$ as a product.

$\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{\sqrt{2}} \cdot \frac{1}{2}$

Step 8.27

Multiply $\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{\sqrt{2}}$ by $\frac{1}{2}$ .

$\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{\sqrt{2} \cdot 2}$

Step 8.28

Move $2$ to the left of $\sqrt{2}$ .

$\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9

Simplify.

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Step 9.1

Simplify each term.

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Step 9.1.1

Multiply $\frac{2}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{25 (2 + \ln (| \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2

Combine and simplify the denominator.

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Step 9.1.2.1

Multiply $\frac{2}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{\sqrt{2} \sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.2

Raise $\sqrt{2}$ to the power of $1$ .

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.3

Raise $\sqrt{2}$ to the power of $1$ .

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.4

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

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{{\sqrt{2}}^{1 + 1}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.5

Add $1$ and $1$ .

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{{\sqrt{2}}^{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 9.1.2.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.6.2

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

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{2^{\frac{1}{2} \cdot 2}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.6.3

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

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{2^{\frac{2}{2}}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.6.4

Cancel the common factor of $2$ .

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Step 9.1.2.6.4.1

Cancel the common factor.

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{2^{\frac{2}{2}}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.6.4.2

Rewrite the expression.

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{2^{1}} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.2.6.5

Evaluate the exponent.

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{2} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.3

Cancel the common factor of $2$ .

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Step 9.1.3.1

Cancel the common factor.

$\frac{25 (2 + \ln (| \frac{2 \sqrt{2}}{2} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.1.3.2

Divide $\sqrt{2}$ by $1$ .

$\frac{25 (2 + \ln (| \sqrt{2} + 1 |) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.2

$\sqrt{2} + 1$ is approximately $2.41421356$ which is positive so remove the absolute value

$\frac{25 (2 + \ln (\sqrt{2} + 1) \sqrt{2} - \ln (| 1 |) \sqrt{2})}{2 \sqrt{2}}$

Step 9.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{25 (2 + \ln (\sqrt{2} + 1) \sqrt{2} - \ln (1) \sqrt{2})}{2 \sqrt{2}}$

Step 9.4

The natural logarithm of $1$ is $0$ .

$\frac{25 (2 + \ln (\sqrt{2} + 1) \sqrt{2} - 0 \sqrt{2})}{2 \sqrt{2}}$

Step 9.5

Multiply $- 1$ by $0$ .

$\frac{25 (2 + \ln (\sqrt{2} + 1) \sqrt{2} + 0 \sqrt{2})}{2 \sqrt{2}}$

Step 9.6

Multiply $0$ by $\sqrt{2}$ .

$\frac{25 (2 + \ln (\sqrt{2} + 1) \sqrt{2} + 0)}{2 \sqrt{2}}$

Step 9.7

Add $2$ and $0$ .

$\frac{25 (\ln (\sqrt{2} + 1) \sqrt{2} + 2)}{2 \sqrt{2}}$

Step 9.8

Apply the distributive property.

$\frac{25 (\ln (\sqrt{2} + 1) \sqrt{2}) + 25 \cdot 2}{2 \sqrt{2}}$

Step 9.9

Multiply $25$ by $2$ .

$\frac{25 \ln (\sqrt{2} + 1) \sqrt{2} + 50}{2 \sqrt{2}}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\frac{25 \ln (\sqrt{2} + 1) \sqrt{2} + 50}{2 \sqrt{2}}$

Decimal Form:

$28.69483936 \dots$

Step 11