Calculus Examples

$\int \sqrt{x^{2} + 10 x} d x$

Step 1

Complete the square.

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Step 1.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 10$

$c = 0$

Step 1.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{10}{2 \cdot 1}$

Step 1.3.2

Cancel the common factor of $10$ and $2$ .

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Step 1.3.2.1

Factor $2$ out of $10$ .

$d = \frac{2 \cdot 5}{2 \cdot 1}$

Step 1.3.2.2

Cancel the common factors.

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Step 1.3.2.2.1

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 5}{2 (1)}$

Step 1.3.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot 5}{2 \cdot 1}$

Step 1.3.2.2.3

Rewrite the expression.

$d = \frac{5}{1}$

Step 1.3.2.2.4

Divide $5$ by $1$ .

$d = 5$

Step 1.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{10^{2}}{4 \cdot 1}$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Raise $10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot 1}$

Step 1.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{100}{4}$

Step 1.4.2.1.3

Divide $100$ by $4$ .

$e = 0 - 1 \cdot 25$

Step 1.4.2.1.4

Multiply $- 1$ by $25$ .

$e = 0 - 25$

Step 1.4.2.2

Subtract $25$ from $0$ .

$e = - 25$

Step 1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 5)}^{2} - 25$ .

$\int \sqrt{{(x + 5)}^{2} - 25} d x$

Step 2

Let $u = x + 5$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = x + 5$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $x + 5$ .

$\frac{d}{d x} [x + 5]$

Step 2.1.2

By the Sum Rule, the derivative of $x + 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 2.1.3

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

$1 + \frac{d}{d x} [5]$

Step 2.1.4

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.1.5

Add $1$ and $0$ .

$1$

Step 2.2

Rewrite the problem using $u$ and $d u$ .

$\int \sqrt{u^{2} - 25} d u$

Step 3

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

$\int \sqrt{{(5 \sec (t))}^{2} - 25} (5 \sec (t) \tan (t)) d t$

Step 4

Simplify terms.

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Step 4.1

Simplify $\sqrt{{(5 \sec (t))}^{2} - 25}$ .

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

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

$\int \sqrt{5^{2} \sec^{2} (t) - 25} (5 \sec (t) \tan (t)) d t$

Step 4.1.1.2

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

$\int \sqrt{25 \sec^{2} (t) - 25} (5 \sec (t) \tan (t)) d t$

Step 4.1.2

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

$\int \sqrt{25 (\sec^{2} (t)) - 25} (5 \sec (t) \tan (t)) d t$

Step 4.1.3

Factor $25$ out of $- 25$ .

$\int \sqrt{25 \sec^{2} (t) + 25 \cdot - 1} (5 \sec (t) \tan (t)) d t$

Step 4.1.4

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

$\int \sqrt{25 (\sec^{2} (t) - 1)} (5 \sec (t) \tan (t)) d t$

Step 4.1.5

Apply pythagorean identity.

$\int \sqrt{25 \tan^{2} (t)} (5 \sec (t) \tan (t)) d t$

Step 4.1.6

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

$\int \sqrt{{(5 \tan (t))}^{2}} (5 \sec (t) \tan (t)) d t$

Step 4.1.7

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

$\int 5 \tan (t) (5 \sec (t) \tan (t)) d t$

Step 4.2

Simplify.

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Step 4.2.1

Multiply $5$ by $5$ .

$\int 25 \tan (t) (\sec (t) \tan (t)) d t$

Step 4.2.2

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

$\int 25 (\tan^{1} (t) \tan (t)) \sec (t) d t$

Step 4.2.3

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

$\int 25 (\tan^{1} (t) \tan^{1} (t)) \sec (t) d t$

Step 4.2.4

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

$\int 25 {\tan (t)}^{1 + 1} \sec (t) d t$

Step 4.2.5

Add $1$ and $1$ .

$\int 25 \tan^{2} (t) \sec (t) d t$

Step 5

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

$25 \int \tan^{2} (t) \sec (t) d t$

Step 6

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

$25 \int \tan^{2} (t) \sec^{1} (t) d t$

Step 7

Using the Pythagorean Identity, rewrite $\tan^{2} (t)$ as $- 1 + \sec^{2} (t)$ .

$25 \int (- 1 + \sec^{2} (t)) \sec^{1} (t) d t$

Step 8

Simplify terms.

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Step 8.1

Apply the distributive property.

$25 \int - 1 \sec^{1} (t) + \sec^{2} (t) \sec^{1} (t) d t$

Step 8.2

Simplify each term.

$25 \int - \sec (t) + \sec^{3} (t) d t$

Step 9

Split the single integral into multiple integrals.

$25 (\int - \sec (t) d t + \int \sec^{3} (t) d t)$

Step 10

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

$25 (- \int \sec (t) d t + \int \sec^{3} (t) d t)$

Step 11

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \int \sec^{3} (t) d t)$

Step 12

Factor $\sec (t)$ out of $\sec^{3} (t)$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \int \sec (t) \sec^{2} (t) d t)$

Step 13

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \sec (t)$ and $d v = \sec^{2} (t)$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \tan (t) (\sec (t) \tan (t)) d t)$

Step 14

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \tan^{1} (t) \tan (t) \sec (t) d t)$

Step 15

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \tan^{1} (t) \tan^{1} (t) \sec (t) d t)$

Step 16

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int {\tan (t)}^{1 + 1} \sec (t) d t)$

Step 17

Simplify the expression.

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Step 17.1

Add $1$ and $1$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \tan^{2} (t) \sec (t) d t)$

Step 17.2

Reorder $\tan^{2} (t)$ and $\sec (t)$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \sec (t) \tan^{2} (t) d t)$

Step 18

Using the Pythagorean Identity, rewrite $\tan^{2} (t)$ as $- 1 + \sec^{2} (t)$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \sec (t) (- 1 + \sec^{2} (t)) d t)$

Step 19

Simplify by multiplying through.

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Step 19.1

Rewrite the exponentiation as a product.

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \sec (t) (- 1 + \sec (t) \sec (t)) d t)$

Step 19.2

Apply the distributive property.

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int \sec (t) \cdot - 1 + \sec (t) (\sec (t) \sec (t)) d t)$

Step 19.3

Reorder $\sec (t)$ and $- 1$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \cdot \sec (t) + \sec (t) (\sec (t) \sec (t)) d t)$

Step 20

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + \sec^{1} (t) \sec (t) \sec (t) d t)$

Step 21

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + \sec^{1} (t) \sec^{1} (t) \sec (t) d t)$

Step 22

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + {\sec (t)}^{1 + 1} \sec (t) d t)$

Step 23

Add $1$ and $1$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + \sec^{2} (t) \sec (t) d t)$

Step 24

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + \sec^{2} (t) \sec^{1} (t) d t)$

Step 25

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + {\sec (t)}^{2 + 1} d t)$

Step 26

Add $2$ and $1$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - \int - 1 \sec (t) + \sec^{3} (t) d t)$

Step 27

Split the single integral into multiple integrals.

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - (\int - 1 \sec (t) d t + \int \sec^{3} (t) d t))$

Step 28

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - (- \int \sec (t) d t + \int \sec^{3} (t) d t))$

Step 29

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

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - (- (\ln (| \sec (t) + \tan (t) |) + C) + \int \sec^{3} (t) d t))$

Step 30

Simplify by multiplying through.

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Step 30.1

Apply the distributive property.

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) - - (\ln (| \sec (t) + \tan (t) |) + C) - \int \sec^{3} (t) d t)$

Step 30.2

Multiply $- 1$ by $- 1$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \sec (t) \tan (t) + 1 (\ln (| \sec (t) + \tan (t) |) + C) - \int \sec^{3} (t) d t)$

Step 31

Solving for $\int \sec^{3} (t) d t$ , we find that $\int \sec^{3} (t) d t$ = $\frac{\sec (t) \tan (t) + 1 (\ln (| \sec (t) + \tan (t) |) + C)}{2}$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \frac{\sec (t) \tan (t) + 1 (\ln (| \sec (t) + \tan (t) |) + C)}{2} + C)$

Step 32

Multiply $\ln (| \sec (t) + \tan (t) |) + C$ by $1$ .

$25 (- (\ln (| \sec (t) + \tan (t) |) + C) + \frac{\sec (t) \tan (t) + \ln (| \sec (t) + \tan (t) |) + C}{2} + C)$

Step 33

Simplify.

$25 \frac{- \ln (| \sec (t) + \tan (t) |) \cdot 2 + \sec (t) \tan (t) + \ln (| \sec (t) + \tan (t) |)}{2} + C$

Step 34

Simplify.

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Step 34.1

Multiply $2$ by $- 1$ .

$25 \frac{- 2 \ln (| \sec (t) + \tan (t) |) + \sec (t) \tan (t) + \ln (| \sec (t) + \tan (t) |)}{2} + C$

Step 34.2

Add $- 2 \ln (| \sec (t) + \tan (t) |)$ and $\ln (| \sec (t) + \tan (t) |)$ .

$25 \frac{\sec (t) \tan (t) - \ln (| \sec (t) + \tan (t) |)}{2} + C$

Step 34.3

Combine $25$ and $\frac{\sec (t) \tan (t) - \ln (| \sec (t) + \tan (t) |)}{2}$ .

$\frac{25 (\sec (t) \tan (t) - \ln (| \sec (t) + \tan (t) |))}{2} + C$

Step 35

Substitute back in for each integration substitution variable.

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Step 35.1

Replace all occurrences of $t$ with $arcsec (\frac{u}{5})$ .

$\frac{25 (\sec (arcsec (\frac{u}{5})) \tan (arcsec (\frac{u}{5})) - \ln (| \sec (arcsec (\frac{u}{5})) + \tan (arcsec (\frac{u}{5})) |))}{2} + C$

Step 35.2

Replace all occurrences of $u$ with $x + 5$ .

$\frac{25 (\sec (arcsec (\frac{x + 5}{5})) \tan (arcsec (\frac{x + 5}{5})) - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36

Simplify.

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Step 36.1

Simplify each term.

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Step 36.1.1

The functions secant and arcsecant are inverses.

$\frac{25 (\frac{x + 5}{5} \tan (arcsec (\frac{x + 5}{5})) - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.2

Draw a triangle in the plane with vertices $(1, \sqrt{{(\frac{x + 5}{5})}^{2} - 1^{2}})$ , $(1, 0)$ , and the origin. Then $arcsec (\frac{x + 5}{5})$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, \sqrt{{(\frac{x + 5}{5})}^{2} - 1^{2}})$ . Therefore, $\tan (arcsec (\frac{x + 5}{5}))$ is $\sqrt{{(\frac{x + 5}{5})}^{2} - 1}$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{{(\frac{x + 5}{5})}^{2} - 1} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.3

Rewrite $1$ as $1^{2}$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{{(\frac{x + 5}{5})}^{2} - 1^{2}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \frac{x + 5}{5}$ and $b = 1$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{(\frac{x + 5}{5} + 1) (\frac{x + 5}{5} - 1)} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5

Simplify.

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Step 36.1.5.1

Write $1$ as a fraction with a common denominator.

$\frac{25 (\frac{x + 5}{5} \sqrt{(\frac{x + 5}{5} + \frac{5}{5}) (\frac{x + 5}{5} - 1)} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.2

Combine the numerators over the common denominator.

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 5 + 5}{5} (\frac{x + 5}{5} - 1)} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.3

Add $5$ and $5$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} (\frac{x + 5}{5} - 1)} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} (\frac{x + 5}{5} - 1 \cdot \frac{5}{5})} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.5

Combine $- 1$ and $\frac{5}{5}$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} (\frac{x + 5}{5} + \frac{- 1 \cdot 5}{5})} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.6

Combine the numerators over the common denominator.

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} \cdot \frac{x + 5 - 1 \cdot 5}{5}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.7

Rewrite $\frac{x + 5 - 1 \cdot 5}{5}$ in a factored form.

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Step 36.1.5.7.1

Multiply $- 1$ by $5$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} \cdot \frac{x + 5 - 5}{5}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.7.2

Subtract $5$ from $5$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} \cdot \frac{x + 0}{5}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.5.7.3

Add $x$ and $0$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{x + 10}{5} \cdot \frac{x}{5}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.6

Multiply $\frac{x + 10}{5}$ by $\frac{x}{5}$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{(x + 10) x}{5 \cdot 5}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.7

Multiply $5$ by $5$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{(x + 10) x}{25}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.8

Rewrite $\frac{(x + 10) x}{25}$ as ${(\frac{1}{5})}^{2} ((x + 10) x)$ .

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Step 36.1.8.1

Factor the perfect power $1^{2}$ out of $(x + 10) x$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{1^{2} ((x + 10) x)}{25}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.8.2

Factor the perfect power $5^{2}$ out of $25$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{\frac{1^{2} ((x + 10) x)}{5^{2} \cdot 1}} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.8.3

Rearrange the fraction $\frac{1^{2} ((x + 10) x)}{5^{2} \cdot 1}$ .

$\frac{25 (\frac{x + 5}{5} \sqrt{{(\frac{1}{5})}^{2} ((x + 10) x)} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.9

Pull terms out from under the radical.

$\frac{25 (\frac{x + 5}{5} (\frac{1}{5} \sqrt{(x + 10) x}) - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.10

Rewrite using the commutative property of multiplication.

$\frac{25 (\frac{1}{5} \cdot \frac{x + 5}{5} \sqrt{(x + 10) x} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.11

Multiply $\frac{1}{5} \cdot \frac{x + 5}{5}$ .

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Step 36.1.11.1

Multiply $\frac{1}{5}$ by $\frac{x + 5}{5}$ .

$\frac{25 (\frac{x + 5}{5 \cdot 5} \sqrt{(x + 10) x} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.11.2

Multiply $5$ by $5$ .

$\frac{25 (\frac{x + 5}{25} \sqrt{(x + 10) x} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.12

Combine $\frac{x + 5}{25}$ and $\sqrt{(x + 10) x}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \sec (arcsec (\frac{x + 5}{5})) + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.13

Simplify each term.

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Step 36.1.13.1

The functions secant and arcsecant are inverses.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \tan (arcsec (\frac{x + 5}{5})) |))}{2} + C$

Step 36.1.13.2

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{{(\frac{x + 5}{5})}^{2} - 1} |))}{2} + C$

Step 36.1.13.3

Rewrite $1$ as $1^{2}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{{(\frac{x + 5}{5})}^{2} - 1^{2}} |))}{2} + C$

Step 36.1.13.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \frac{x + 5}{5}$ and $b = 1$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{(\frac{x + 5}{5} + 1) (\frac{x + 5}{5} - 1)} |))}{2} + C$

Step 36.1.13.5

Simplify.

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Step 36.1.13.5.1

Write $1$ as a fraction with a common denominator.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{(\frac{x + 5}{5} + \frac{5}{5}) (\frac{x + 5}{5} - 1)} |))}{2} + C$

Step 36.1.13.5.2

Combine the numerators over the common denominator.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 5 + 5}{5} (\frac{x + 5}{5} - 1)} |))}{2} + C$

Step 36.1.13.5.3

Add $5$ and $5$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} (\frac{x + 5}{5} - 1)} |))}{2} + C$

Step 36.1.13.5.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} (\frac{x + 5}{5} - 1 \cdot \frac{5}{5})} |))}{2} + C$

Step 36.1.13.5.5

Combine $- 1$ and $\frac{5}{5}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} (\frac{x + 5}{5} + \frac{- 1 \cdot 5}{5})} |))}{2} + C$

Step 36.1.13.5.6

Combine the numerators over the common denominator.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} \cdot \frac{x + 5 - 1 \cdot 5}{5}} |))}{2} + C$

Step 36.1.13.5.7

Rewrite $\frac{x + 5 - 1 \cdot 5}{5}$ in a factored form.

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Step 36.1.13.5.7.1

Multiply $- 1$ by $5$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} \cdot \frac{x + 5 - 5}{5}} |))}{2} + C$

Step 36.1.13.5.7.2

Subtract $5$ from $5$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} \cdot \frac{x + 0}{5}} |))}{2} + C$

Step 36.1.13.5.7.3

Add $x$ and $0$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{x + 10}{5} \cdot \frac{x}{5}} |))}{2} + C$

Step 36.1.13.6

Multiply $\frac{x + 10}{5}$ by $\frac{x}{5}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{(x + 10) x}{5 \cdot 5}} |))}{2} + C$

Step 36.1.13.7

Multiply $5$ by $5$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{(x + 10) x}{25}} |))}{2} + C$

Step 36.1.13.8

Rewrite $\frac{(x + 10) x}{25}$ as ${(\frac{1}{5})}^{2} ((x + 10) x)$ .

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Step 36.1.13.8.1

Factor the perfect power $1^{2}$ out of $(x + 10) x$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{1^{2} ((x + 10) x)}{25}} |))}{2} + C$

Step 36.1.13.8.2

Factor the perfect power $5^{2}$ out of $25$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{\frac{1^{2} ((x + 10) x)}{5^{2} \cdot 1}} |))}{2} + C$

Step 36.1.13.8.3

Rearrange the fraction $\frac{1^{2} ((x + 10) x)}{5^{2} \cdot 1}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \sqrt{{(\frac{1}{5})}^{2} ((x + 10) x)} |))}{2} + C$

Step 36.1.13.9

Pull terms out from under the radical.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \frac{1}{5} \sqrt{(x + 10) x} |))}{2} + C$

Step 36.1.13.10

Combine $\frac{1}{5}$ and $\sqrt{(x + 10) x}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5}{5} + \frac{\sqrt{(x + 10) x}}{5} |))}{2} + C$

Step 36.1.14

Combine the numerators over the common denominator.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5 + \sqrt{(x + 10) x}}{5} |))}{2} + C$

Step 36.1.15

Reorder factors in $x + 5 + \sqrt{(x + 10) x}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (| \frac{x + 5 + \sqrt{x (x + 10)}}{5} |))}{2} + C$

Step 36.1.16

Remove non-negative terms from the absolute value.

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5}))}{2} + C$

Step 36.2

To write $- \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5})$ as a fraction with a common denominator, multiply by $\frac{25}{25}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} - \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5}) \cdot \frac{25}{25})}{2} + C$

Step 36.3

Combine $- \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5})$ and $\frac{25}{25}$ .

$\frac{25 (\frac{(x + 5) \sqrt{(x + 10) x}}{25} + \frac{- \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5}) \cdot 25}{25})}{2} + C$

Step 36.4

Combine the numerators over the common denominator.

$\frac{25 \frac{(x + 5) \sqrt{(x + 10) x} - \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5}) \cdot 25}{25}}{2} + C$

Step 36.5

Cancel the common factor of $25$ .

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Step 36.5.1

Cancel the common factor.

$\frac{25 \frac{(x + 5) \sqrt{(x + 10) x} - \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5}) \cdot 25}{25}}{2} + C$

Step 36.5.2

Rewrite the expression.

$\frac{(x + 5) \sqrt{(x + 10) x} - \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5}) \cdot 25}{2} + C$

Step 36.6

Multiply $25$ by $- 1$ .

$\frac{(x + 5) \sqrt{(x + 10) x} - 25 \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5})}{2} + C$

Step 36.7

Reorder factors in $(x + 5) \sqrt{(x + 10) x} - 25 \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5})$ .

$\frac{\sqrt{x (x + 10)} (x + 5) - 25 \ln (\frac{| x + 5 + \sqrt{x (x + 10)} |}{5})}{2} + C$

Step 37

Reorder terms.

$\frac{1}{2} (\sqrt{x (x + 10)} (x + 5) - 25 \ln (\frac{1}{5} | x + 5 + \sqrt{x (x + 10)} |)) + C$