Calculus Examples

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

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 $c$ .

$a = 1$

$b = 2$

$c = 0$

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 $d = \frac{b}{2 a}$ .

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

Step 1.3.2

Cancel the common factor of $2$ .

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

$d = 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

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

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 $2$ to the power of $2$ .

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

Step 1.4.2.1.2

Multiply $4$ by $1$ .

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

Step 1.4.2.1.3

Cancel the common factor of $4$ .

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Step 1.4.2.1.3.1

Cancel the common factor.

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

Step 1.4.2.1.3.2

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Step 1.4.2.1.4

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Step 1.4.2.2

Subtract $1$ from $0$ .

$e = - 1$

Step 1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 1)}^{2} - 1$ .

$\int \sqrt{{(x + 1)}^{2} - 1} d x$

Step 2

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $x + 1$ .

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

Step 2.1.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

Step 2.1.3

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

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

Step 2.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ 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 $d u$ .

$\int \sqrt{u^{2} - 1} d u$

Step 3

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

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

Step 4

Simplify terms.

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Step 4.1

Simplify $\sqrt{\sec^{2} (t) - 1}$ .

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Step 4.1.1

Apply pythagorean identity.

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

Step 4.1.2

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

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

Step 4.2

Simplify.

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Step 4.2.1

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

Add $1$ and $1$ .

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

Step 5

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

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

Step 6

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

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

Step 7

Simplify terms.

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Step 7.1

Apply the distributive property.

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

Step 7.2

Simplify each term.

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

Step 8

Split the single integral into multiple integrals.

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

Step 9

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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

Step 13

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

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

Step 14

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

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

Step 15

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

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

Step 16

Simplify the expression.

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Step 16.1

Add $1$ and $1$ .

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

Step 16.2

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

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

Step 17

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

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

Step 18

Simplify by multiplying through.

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Step 18.1

Rewrite the exponentiation as a product.

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

Step 18.2

Apply the distributive property.

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

Step 18.3

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

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

Step 19

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

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

Step 20

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

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

Step 21

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

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

Step 22

Add $1$ and $1$ .

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

Step 23

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

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

Step 24

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

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

Step 25

Add $2$ and $1$ .

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

Step 26

Split the single integral into multiple integrals.

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

Step 27

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

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

Step 28

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

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

Step 29

Simplify by multiplying through.

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Step 29.1

Apply the distributive property.

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

Step 29.2

Multiply $- 1$ by $- 1$ .

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

Step 30

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

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

Step 31

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

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

Step 32

Simplify.

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

Step 33

Substitute back in for each integration substitution variable.

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Step 33.1

Replace all occurrences of $t$ with $arcsec (u)$ .

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

Step 33.2

Replace all occurrences of $u$ with $x + 1$ .

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