Calculus Examples

$\int_{- 1}^{1} \frac{3 x^{2} + 2 x + 1}{x + 2} d x$

Step 1

Divide

$3 x^{2} + 2 x + 1$ by

$x + 2$ .

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of

$0$ .

$x$

$2$

$3 x^{2}$

$2 x$

$1$

Step 1.2

Divide the highest order term in the dividend

$3 x^{2}$ by the highest order term in divisor

$x$ .

					$3 x$
	$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$

Step 1.3

Multiply the new quotient term by the divisor.

				$3 x$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			+	$3 x^{2}$	+	$6 x$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in

$3 x^{2} + 6 x$

				$3 x$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$3 x$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$
					-	$4 x$

Step 1.6

Pull the next terms from the original dividend down into the current dividend.

				$3 x$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$
					-	$4 x$	+	$1$

Step 1.7

Divide the highest order term in the dividend

$- 4 x$ by the highest order term in divisor

$x$ .

				$3 x$	-	$4$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$
					-	$4 x$	+	$1$

Step 1.8

Multiply the new quotient term by the divisor.

				$3 x$	-	$4$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$
					-	$4 x$	+	$1$
					-	$4 x$	-	$8$

Step 1.9

The expression needs to be subtracted from the dividend, so change all the signs in

$- 4 x - 8$

				$3 x$	-	$4$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$
					-	$4 x$	+	$1$
					+	$4 x$	+	$8$

Step 1.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$3 x$	-	$4$
$x$	+	$2$		$3 x^{2}$	+	$2 x$	+	$1$
			-	$3 x^{2}$	-	$6 x$
					-	$4 x$	+	$1$
					+	$4 x$	+	$8$
							+	$9$

Step 1.11

The final answer is the quotient plus the remainder over the divisor.

$\int_{- 1}^{1} 3 x - 4 + \frac{9}{x + 2} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{- 1}^{1} 3 x d x + \int_{- 1}^{1} - 4 d x + \int_{- 1}^{1} \frac{9}{x + 2} d x$

Step 3

Since

$3$ is constant with respect to

$x$ , move

$3$ out of the integral.

$3 \int_{- 1}^{1} x d x + \int_{- 1}^{1} - 4 d x + \int_{- 1}^{1} \frac{9}{x + 2} d x$

Step 4

By the Power Rule, the integral of

$x$ with respect to

$x$ is

$\frac{1}{2} x^{2}$ .

$3 (\frac{1}{2} x^{2}]_{- 1}^{1}) + \int_{- 1}^{1} - 4 d x + \int_{- 1}^{1} \frac{9}{x + 2} d x$

Step 5

Combine

$\frac{1}{2}$ and

$x^{2}$ .

$3 (\frac{x^{2}}{2}]_{- 1}^{1}) + \int_{- 1}^{1} - 4 d x + \int_{- 1}^{1} \frac{9}{x + 2} d x$

Step 6

Apply the constant rule.

$3 (\frac{x^{2}}{2}]_{- 1}^{1}) + - 4 x]_{- 1}^{1} + \int_{- 1}^{1} \frac{9}{x + 2} d x$

Step 7

Since

$9$ is constant with respect to

$x$ , move

$9$ out of the integral.

$3 (\frac{x^{2}}{2}]_{- 1}^{1}) + - 4 x]_{- 1}^{1} + 9 \int_{- 1}^{1} \frac{1}{x + 2} d x$

Step 8

Let

$u = x + 2$ . Then

$d u = d x$ . Rewrite using

$u$ and

$d$

$u$ .

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Step 8.1

Let

$u = x + 2$ . Find

$\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate

$x + 2$ .

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

Step 8.1.2

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 8.1.3

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 8.1.4

Since

$2$ is constant with respect to

$x$ , the derivative of

$2$ with respect to

$x$ is

$0$ .

$1 + 0$

Step 8.1.5

Add

$1$ and

$0$ .

$1$

Step 8.2

Substitute the lower limit in for

$x$ in

$u = x + 2$ .

$u_{lower} = - 1 + 2$

Step 8.3

Add

$- 1$ and

$2$ .

$u_{lower} = 1$

Step 8.4

Substitute the upper limit in for

$x$ in

$u = x + 2$ .

$u_{upper} = 1 + 2$

Step 8.5

Add

$1$ and

$2$ .

$u_{upper} = 3$

Step 8.6

The values found for

$u_{lower}$ and

$u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 3$

Step 8.7

Rewrite the problem using

$u$ ,

$d u$ , and the new limits of integration.

$3 (\frac{x^{2}}{2}]_{- 1}^{1}) + - 4 x]_{- 1}^{1} + 9 \int_{1}^{3} \frac{1}{u} d u$

Step 9

The integral of

$\frac{1}{u}$ with respect to

$u$ is

$\ln (| u |)$ .

$3 (\frac{x^{2}}{2}]_{- 1}^{1}) + - 4 x]_{- 1}^{1} + 9 (\ln (| u |)]_{1}^{3})$

Step 10

Substitute and simplify.

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Step 10.1

Evaluate

$\frac{x^{2}}{2}$ at

$1$ and at

$- 1$ .

$3 ((\frac{1^{2}}{2}) - \frac{{(- 1)}^{2}}{2}) + - 4 x]_{- 1}^{1} + 9 (\ln (| u |)]_{1}^{3})$

Step 10.2

Evaluate

$- 4 x$ at

$1$ and at

$- 1$ .

$3 (\frac{1^{2}}{2} - \frac{{(- 1)}^{2}}{2}) + - 4 \cdot 1 + 4 \cdot - 1 + 9 (\ln (| u |)]_{1}^{3})$

Step 10.3

Evaluate

$\ln (| u |)$ at

$3$ and at

$1$ .

$3 (\frac{1^{2}}{2} - \frac{{(- 1)}^{2}}{2}) + - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4

Simplify.

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Step 10.4.1

One to any power is one.

$3 (\frac{1}{2} - \frac{{(- 1)}^{2}}{2}) - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.2

Raise

$- 1$ to the power of

$2$ .

$3 (\frac{1}{2} - \frac{1}{2}) - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.3

Combine the numerators over the common denominator.

$3 \frac{1 - 1}{2} - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.4

Subtract

$1$ from

$1$ .

$3 (\frac{0}{2}) - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.5

Cancel the common factor of

$0$ and

$2$ .

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Step 10.4.5.1

Factor

$2$ out of

$0$ .

$3 \frac{2 (0)}{2} - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.5.2

Cancel the common factors.

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Step 10.4.5.2.1

Factor

$2$ out of

$2$ .

$3 \frac{2 \cdot 0}{2 \cdot 1} - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.5.2.2

Cancel the common factor.

$3 \frac{2 \cdot 0}{2 \cdot 1} - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.5.2.3

Rewrite the expression.

$3 (\frac{0}{1}) - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.5.2.4

Divide

$0$ by

$1$ .

$3 \cdot 0 - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.6

Multiply

$3$ by

$0$ .

$0 - 4 \cdot 1 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.7

Multiply

$- 4$ by

$1$ .

$0 - 4 + 4 \cdot - 1 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.8

Multiply

$4$ by

$- 1$ .

$0 - 4 - 4 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.9

Subtract

$4$ from

$- 4$ .

$0 - 8 + 9 ((\ln (| 3 |)) - \ln (| 1 |))$

Step 10.4.10

Subtract

$8$ from

$0$ .

$- 8 + 9 (\ln (| 3 |) - \ln (| 1 |))$

Step 11

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$- 8 + 9 \ln (\frac{| 3 |}{| 1 |})$

Step 12

Simplify.

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Step 12.1

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

$0$ and

$3$ is

$3$ .

$- 8 + 9 \ln (\frac{3}{| 1 |})$

Step 12.2

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

$0$ and

$1$ is

$1$ .

$- 8 + 9 \ln (\frac{3}{1})$

Step 12.3

Divide

$3$ by

$1$ .

$- 8 + 9 \ln (3)$

Step 13

The result can be shown in multiple forms.

Exact Form:

$- 8 + 9 \ln (3)$

Decimal Form:

$1.88751059 \dots$

Step 14