Calculus Examples

$\int \frac{8}{\sqrt{12 - x^{2} - 4 x}} d x$

Step 1

Factor by grouping.

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Step 1.1

Reorder terms.

$\int \frac{8}{\sqrt{- x^{2} - 4 x + 12}} d x$

Step 1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 12 = - 12$ and whose sum is $b = - 4$ .

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Step 1.2.1

Factor $- 4$ out of $- 4 x$ .

$\int \frac{8}{\sqrt{- x^{2} - 4 (x) + 12}} d x$

Step 1.2.2

Rewrite $- 4$ as $2$ plus $- 6$

$\int \frac{8}{\sqrt{- x^{2} + (2 - 6) x + 12}} d x$

Step 1.2.3

Apply the distributive property.

$\int \frac{8}{\sqrt{- x^{2} + 2 x - 6 x + 12}} d x$

Step 1.3

Factor out the greatest common factor from each group.

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Step 1.3.1

Group the first two terms and the last two terms.

$\int \frac{8}{\sqrt{(- x^{2} + 2 x) - 6 x + 12}} d x$

Step 1.3.2

Factor out the greatest common factor (GCF) from each group.

$\int \frac{8}{\sqrt{x (- x + 2) + 6 (- x + 2)}} d x$

Step 1.4

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

$\int \frac{8}{\sqrt{(- x + 2) (x + 6)}} d x$

Step 2

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$8 \int \frac{1}{\sqrt{(- x + 2) (x + 6)}} d x$

Step 3

Complete the square.

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Step 3.1

Simplify the expression.

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Step 3.1.1

Expand $(- x + 2) (x + 6)$ using the FOIL Method.

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Step 3.1.1.1

Apply the distributive property.

$- x (x + 6) + 2 (x + 6)$

Step 3.1.1.2

Apply the distributive property.

$- x \cdot x - x \cdot 6 + 2 (x + 6)$

Step 3.1.1.3

Apply the distributive property.

$- x \cdot x - x \cdot 6 + 2 x + 2 \cdot 6$

Step 3.1.2

Simplify and combine like terms.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 3.1.2.1.1.1

Move $x$ .

$- (x \cdot x) - x \cdot 6 + 2 x + 2 \cdot 6$

Step 3.1.2.1.1.2

Multiply $x$ by $x$ .

$- x^{2} - x \cdot 6 + 2 x + 2 \cdot 6$

Step 3.1.2.1.2

Multiply $6$ by $- 1$ .

$- x^{2} - 6 x + 2 x + 2 \cdot 6$

Step 3.1.2.1.3

Multiply $2$ by $6$ .

$- x^{2} - 6 x + 2 x + 12$

Step 3.1.2.2

Add $- 6 x$ and $2 x$ .

$- x^{2} - 4 x + 12$

Step 3.2

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

$a = - 1$

$b = - 4$

$c = 12$

Step 3.3

Consider the vertex form of a parabola.

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

Step 3.4

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

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Step 3.4.1

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

$d = \frac{- 4}{2 \cdot - 1}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Cancel the common factor of $- 4$ and $2$ .

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Step 3.4.2.1.1

Factor $2$ out of $- 4$ .

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

Step 3.4.2.1.2

Move the negative one from the denominator of $\frac{- 2}{- 1}$ .

$d = - 1 \cdot - 2$

Step 3.4.2.2

Rewrite $- 1 \cdot - 2$ as $- - 2$ .

$d = - - 2$

Step 3.4.2.3

Multiply $- 1$ by $- 2$ .

$d = 2$

Step 3.5

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

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Step 3.5.1

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

$e = 12 - \frac{{(- 4)}^{2}}{4 \cdot - 1}$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

Cancel the common factor of ${(- 4)}^{2}$ and $4$ .

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Step 3.5.2.1.1.1

Rewrite $- 4$ as $- 1 (4)$ .

$e = 12 - \frac{{(- 1 (4))}^{2}}{4 \cdot - 1}$

Step 3.5.2.1.1.2

Apply the product rule to $- 1 (4)$ .

$e = 12 - \frac{{(- 1)}^{2} \cdot 4^{2}}{4 \cdot - 1}$

Step 3.5.2.1.1.3

Raise $- 1$ to the power of $2$ .

$e = 12 - \frac{1 \cdot 4^{2}}{4 \cdot - 1}$

Step 3.5.2.1.1.4

Multiply $4^{2}$ by $1$ .

$e = 12 - \frac{4^{2}}{4 \cdot - 1}$

Step 3.5.2.1.1.5

Factor $4$ out of $4^{2}$ .

$e = 12 - \frac{4 \cdot 4}{4 \cdot - 1}$

Step 3.5.2.1.1.6

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$e = 12 - (- 1 \cdot 4)$

Step 3.5.2.1.2

Multiply $- (- 1 \cdot 4)$ .

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Step 3.5.2.1.2.1

Multiply $- 1$ by $4$ .

$e = 12 - - 4$

Step 3.5.2.1.2.2

Multiply $- 1$ by $- 4$ .

$e = 12 + 4$

Step 3.5.2.2

Add $12$ and $4$ .

$e = 16$

Step 3.6

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 2)}^{2} + 16$ .

$8 \int \frac{1}{\sqrt{- {(x + 2)}^{2} + 16}} d x$

Step 4

Let $u = x + 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = x + 2$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $x + 2$ .

$\frac{d}{d x} [x + 2]$

Step 4.1.2

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 4.1.3

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

$1 + \frac{d}{d x} [2]$

Step 4.1.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$1 + 0$

Step 4.1.5

Add $1$ and $0$ .

$1$

Step 4.2

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

$8 \int \frac{1}{\sqrt{- u^{2} + 16}} d u$

Step 5

Simplify the expression.

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Step 5.1

Rewrite $16$ as $4^{2}$ .

$8 \int \frac{1}{\sqrt{- u^{2} + 4^{2}}} d u$

Step 5.2

Reorder $- u^{2}$ and $4^{2}$ .

$8 \int \frac{1}{\sqrt{4^{2} - u^{2}}} d u$

Step 6

The integral of $\frac{1}{\sqrt{4^{2} - u^{2}}}$ with respect to $u$ is $\arcsin (\frac{u}{4})$

$8 (\arcsin (\frac{u}{4}) + C)$

Step 7

Rewrite $8 (\arcsin (\frac{u}{4}) + C)$ as $8 \arcsin (\frac{1}{4} u) + C$ .

$8 \arcsin (\frac{1}{4} u) + C$

Step 8

Replace all occurrences of $u$ with $x + 2$ .

$8 \arcsin (\frac{1}{4} (x + 2)) + C$

Step 9

Simplify.

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Step 9.1

Apply the distributive property.

$8 \arcsin (\frac{1}{4} x + \frac{1}{4} \cdot 2) + C$

Step 9.2

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

$8 \arcsin (\frac{x}{4} + \frac{1}{4} \cdot 2) + C$

Step 9.3

Cancel the common factor of $2$ .

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Step 9.3.1

Factor $2$ out of $4$ .

$8 \arcsin (\frac{x}{4} + \frac{1}{2 (2)} \cdot 2) + C$

Step 9.3.2

Cancel the common factor.

$8 \arcsin (\frac{x}{4} + \frac{1}{2 \cdot 2} \cdot 2) + C$

Step 9.3.3

Rewrite the expression.

$8 \arcsin (\frac{x}{4} + \frac{1}{2}) + C$

Step 10

Reorder terms.

$8 \arcsin (\frac{1}{4} x + \frac{1}{2}) + C$