Calculus Examples

$\int_{1}^{3} \frac{t^{2}}{5 - 2 t} d t$

Step 1

Reorder $5$ and $- 2 t$ .

$\int_{1}^{3} \frac{t^{2}}{- 2 t + 5} d t$

Step 2

Divide $t^{2}$ by $- 2 t + 5$ .

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Step 2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$2 t$

$5$

$t^{2}$

$0 t$

$0$

Step 2.2

Divide the highest order term in the dividend $t^{2}$ by the highest order term in divisor $- 2 t$ .

				-	$\frac{t}{2}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$

Step 2.3

Multiply the new quotient term by the divisor.

				-	$\frac{t}{2}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				+	$t^{2}$	-	$\frac{5 t}{2}$

Step 2.4

The expression needs to be subtracted from the dividend, so change all the signs in $t^{2} - \frac{5 t}{2}$

				-	$\frac{t}{2}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$

Step 2.5

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

				-	$\frac{t}{2}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$
						+	$\frac{5 t}{2}$

Step 2.6

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

				-	$\frac{t}{2}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$
						+	$\frac{5 t}{2}$	+	$0$

Step 2.7

Divide the highest order term in the dividend $\frac{5 t}{2}$ by the highest order term in divisor $- 2 t$ .

				-	$\frac{t}{2}$	-	$\frac{5}{4}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$
						+	$\frac{5 t}{2}$	+	$0$

Step 2.8

Multiply the new quotient term by the divisor.

				-	$\frac{t}{2}$	-	$\frac{5}{4}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$
						+	$\frac{5 t}{2}$	+	$0$
						+	$\frac{5 t}{2}$	-	$\frac{25}{4}$

Step 2.9

The expression needs to be subtracted from the dividend, so change all the signs in $\frac{5 t}{2} - \frac{25}{4}$

				-	$\frac{t}{2}$	-	$\frac{5}{4}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$
						+	$\frac{5 t}{2}$	+	$0$
						-	$\frac{5 t}{2}$	+	$\frac{25}{4}$

Step 2.10

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

				-	$\frac{t}{2}$	-	$\frac{5}{4}$
-	$2 t$	+	$5$		$t^{2}$	+	$0 t$	+	$0$
				-	$t^{2}$	+	$\frac{5 t}{2}$
						+	$\frac{5 t}{2}$	+	$0$
						-	$\frac{5 t}{2}$	+	$\frac{25}{4}$
								+	$\frac{25}{4}$

Step 2.11

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

$\int_{1}^{3} - \frac{t}{2} - \frac{5}{4} + \frac{25}{4 (- 2 t + 5)} d t$

Step 3

Split the single integral into multiple integrals.

$\int_{1}^{3} - \frac{t}{2} d t + \int_{1}^{3} - \frac{5}{4} d t + \int_{1}^{3} \frac{25}{4 (- 2 t + 5)} d t$

Step 4

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

$- \int_{1}^{3} \frac{t}{2} d t + \int_{1}^{3} - \frac{5}{4} d t + \int_{1}^{3} \frac{25}{4 (- 2 t + 5)} d t$

Step 5

Since $\frac{1}{2}$ is constant with respect to $t$ , move $\frac{1}{2}$ out of the integral.

$- (\frac{1}{2} \int_{1}^{3} t d t) + \int_{1}^{3} - \frac{5}{4} d t + \int_{1}^{3} \frac{25}{4 (- 2 t + 5)} d t$

Step 6

By the Power Rule, the integral of $t$ with respect to $t$ is $\frac{1}{2} t^{2}$ .

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + \int_{1}^{3} - \frac{5}{4} d t + \int_{1}^{3} \frac{25}{4 (- 2 t + 5)} d t$

Step 7

Apply the constant rule.

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \int_{1}^{3} \frac{25}{4 (- 2 t + 5)} d t$

Step 8

Since $\frac{25}{4}$ is constant with respect to $t$ , move $\frac{25}{4}$ out of the integral.

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} \int_{1}^{3} \frac{1}{- 2 t + 5} d t$

Step 9

Let $u = - 2 t + 5$ . Then $d u = - 2 d t$ , so $- \frac{1}{2} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 9.1

Let $u = - 2 t + 5$ . Find $\frac{d u}{d t}$ .

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Step 9.1.1

Differentiate $- 2 t + 5$ .

$\frac{d}{d t} [- 2 t + 5]$

Step 9.1.2

By the Sum Rule, the derivative of $- 2 t + 5$ with respect to $t$ is $\frac{d}{d t} [- 2 t] + \frac{d}{d t} [5]$ .

$\frac{d}{d t} [- 2 t] + \frac{d}{d t} [5]$

Step 9.1.3

Evaluate $\frac{d}{d t} [- 2 t]$ .

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Step 9.1.3.1

Since $- 2$ is constant with respect to $t$ , the derivative of $- 2 t$ with respect to $t$ is $- 2 \frac{d}{d t} [t]$ .

$- 2 \frac{d}{d t} [t] + \frac{d}{d t} [5]$

Step 9.1.3.2

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

$- 2 \cdot 1 + \frac{d}{d t} [5]$

Step 9.1.3.3

Multiply $- 2$ by $1$ .

$- 2 + \frac{d}{d t} [5]$

Step 9.1.4

Differentiate using the Constant Rule.

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Step 9.1.4.1

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

$- 2 + 0$

Step 9.1.4.2

Add $- 2$ and $0$ .

$- 2$

Step 9.2

Substitute the lower limit in for $t$ in $u = - 2 t + 5$ .

$u_{lower} = - 2 \cdot 1 + 5$

Step 9.3

Simplify.

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Step 9.3.1

Multiply $- 2$ by $1$ .

$u_{lower} = - 2 + 5$

Step 9.3.2

Add $- 2$ and $5$ .

$u_{lower} = 3$

Step 9.4

Substitute the upper limit in for $t$ in $u = - 2 t + 5$ .

$u_{upper} = - 2 \cdot 3 + 5$

Step 9.5

Simplify.

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Step 9.5.1

Multiply $- 2$ by $3$ .

$u_{upper} = - 6 + 5$

Step 9.5.2

Add $- 6$ and $5$ .

$u_{upper} = - 1$

Step 9.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 3$

$u_{upper} = - 1$

Step 9.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} \int_{3}^{- 1} \frac{1}{u} \cdot \frac{1}{- 2} d u$

Step 10

Simplify.

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Step 10.1

Move the negative in front of the fraction.

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} \int_{3}^{- 1} \frac{1}{u} (- \frac{1}{2}) d u$

Step 10.2

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

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} \int_{3}^{- 1} - \frac{1}{u \cdot 2} d u$

Step 10.3

Move $2$ to the left of $u$ .

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} \int_{3}^{- 1} - \frac{1}{2 u} d u$

Step 11

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

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} (- \int_{3}^{- 1} \frac{1}{2 u} d u)$

Step 12

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} + \frac{25}{4} (- (\frac{1}{2} \int_{3}^{- 1} \frac{1}{u} d u))$

Step 13

Simplify.

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Step 13.1

Multiply $\frac{25}{4}$ by $\frac{1}{2}$ .

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} - \frac{25}{4 \cdot 2} \int_{3}^{- 1} \frac{1}{u} d u$

Step 13.2

Multiply $4$ by $2$ .

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} - \frac{25}{8} \int_{3}^{- 1} \frac{1}{u} d u$

Step 14

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$- \frac{1}{2} \frac{1}{2} t^{2}]_{1}^{3} + - \frac{5}{4} t]_{1}^{3} - \frac{25}{8} \ln (| u |)]_{3}^{- 1}$

Step 15

Substitute and simplify.

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Step 15.1

Evaluate $\frac{1}{2} t^{2}$ at $3$ and at $1$ .

$- \frac{1}{2} ((\frac{1}{2} \cdot 3^{2}) - \frac{1}{2} \cdot 1^{2}) + - \frac{5}{4} t]_{1}^{3} - \frac{25}{8} \ln (| u |)]_{3}^{- 1}$

Step 15.2

Evaluate $- \frac{5}{4} t$ at $3$ and at $1$ .

$- \frac{1}{2} ((\frac{1}{2} \cdot 3^{2}) - \frac{1}{2} \cdot 1^{2}) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} \ln (| u |)]_{3}^{- 1}$

Step 15.3

Evaluate $\ln (| u |)$ at $- 1$ and at $3$ .

$- \frac{1}{2} ((\frac{1}{2} \cdot 3^{2}) - \frac{1}{2} \cdot 1^{2}) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4

Simplify.

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Step 15.4.1

Raise $3$ to the power of $2$ .

$- \frac{1}{2} (\frac{1}{2} \cdot 9 - \frac{1}{2} \cdot 1^{2}) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.2

Combine $\frac{1}{2}$ and $9$ .

$- \frac{1}{2} (\frac{9}{2} - \frac{1}{2} \cdot 1^{2}) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.3

One to any power is one.

$- \frac{1}{2} (\frac{9}{2} - \frac{1}{2} \cdot 1) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.4

Multiply $- 1$ by $1$ .

$- \frac{1}{2} (\frac{9}{2} - \frac{1}{2}) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.5

Combine the numerators over the common denominator.

$- \frac{1}{2} \cdot \frac{9 - 1}{2} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.6

Subtract $1$ from $9$ .

$- \frac{1}{2} \cdot \frac{8}{2} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.7

Cancel the common factor of $8$ and $2$ .

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Step 15.4.7.1

Factor $2$ out of $8$ .

$- \frac{1}{2} \cdot \frac{2 \cdot 4}{2} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.7.2

Cancel the common factors.

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Step 15.4.7.2.1

Factor $2$ out of $2$ .

$- \frac{1}{2} \cdot \frac{2 \cdot 4}{2 (1)} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.7.2.2

Cancel the common factor.

$- \frac{1}{2} \cdot \frac{2 \cdot 4}{2 \cdot 1} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.7.2.3

Rewrite the expression.

$- \frac{1}{2} \cdot \frac{4}{1} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.7.2.4

Divide $4$ by $1$ .

$- \frac{1}{2} \cdot 4 + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.8

Multiply $4$ by $- 1$ .

$- 4 (\frac{1}{2}) + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.9

Combine $- 4$ and $\frac{1}{2}$ .

$\frac{- 4}{2} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.10

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

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Step 15.4.10.1

Factor $2$ out of $- 4$ .

$\frac{2 \cdot - 2}{2} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.10.2

Cancel the common factors.

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Step 15.4.10.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot - 2}{2 (1)} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.10.2.2

Cancel the common factor.

$\frac{2 \cdot - 2}{2 \cdot 1} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.10.2.3

Rewrite the expression.

$\frac{- 2}{1} + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.10.2.4

Divide $- 2$ by $1$ .

$- 2 + (- \frac{5}{4} \cdot 3) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.11

Multiply $3$ by $- 1$ .

$- 2 - 3 (\frac{5}{4}) + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.12

Combine $- 3$ and $\frac{5}{4}$ .

$- 2 + \frac{- 3 \cdot 5}{4} + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.13

Multiply $- 3$ by $5$ .

$- 2 + \frac{- 15}{4} + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.14

Move the negative in front of the fraction.

$- 2 - \frac{15}{4} + \frac{5}{4} \cdot 1 - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.15

Multiply $\frac{5}{4}$ by $1$ .

$- 2 - \frac{15}{4} + \frac{5}{4} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.16

Combine the numerators over the common denominator.

$- 2 + \frac{- 15 + 5}{4} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.17

Add $- 15$ and $5$ .

$- 2 + \frac{- 10}{4} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.18

Cancel the common factor of $- 10$ and $4$ .

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Step 15.4.18.1

Factor $2$ out of $- 10$ .

$- 2 + \frac{2 (- 5)}{4} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.18.2

Cancel the common factors.

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Step 15.4.18.2.1

Factor $2$ out of $4$ .

$- 2 + \frac{2 \cdot - 5}{2 \cdot 2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.18.2.2

Cancel the common factor.

$- 2 + \frac{2 \cdot - 5}{2 \cdot 2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.18.2.3

Rewrite the expression.

$- 2 + \frac{- 5}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.19

Move the negative in front of the fraction.

$- 2 - \frac{5}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.20

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 2 \cdot \frac{2}{2} - \frac{5}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.21

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

$\frac{- 2 \cdot 2}{2} - \frac{5}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.22

Combine the numerators over the common denominator.

$\frac{- 2 \cdot 2 - 5}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.23

Simplify the numerator.

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Step 15.4.23.1

Multiply $- 2$ by $2$ .

$\frac{- 4 - 5}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.23.2

Subtract $5$ from $- 4$ .

$\frac{- 9}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.24

Move the negative in front of the fraction.

$- \frac{9}{2} - \frac{25}{8} ((\ln (| - 1 |)) - \ln (| 3 |))$

Step 15.4.25

To write $- \frac{25}{8} (\ln (| - 1 |) - \ln (| 3 |))$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \frac{9}{2} - \frac{25}{8} (\ln (| - 1 |) - \ln (| 3 |)) \cdot \frac{2}{2}$

Step 15.4.26

Combine $- \frac{25}{8} (\ln (| - 1 |) - \ln (| 3 |))$ and $\frac{2}{2}$ .

$- \frac{9}{2} + \frac{- \frac{25}{8} (\ln (| - 1 |) - \ln (| 3 |)) \cdot 2}{2}$

Step 15.4.27

Combine the numerators over the common denominator.

$\frac{- 9 - \frac{25}{8} (\ln (| - 1 |) - \ln (| 3 |)) \cdot 2}{2}$

Step 15.4.28

Multiply $2$ by $- 1$ .

$\frac{- 9 - 2 (\frac{25}{8}) (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.29

Combine $- 2$ and $\frac{25}{8}$ .

$\frac{- 9 + \frac{- 2 \cdot 25}{8} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.30

Multiply $- 2$ by $25$ .

$\frac{- 9 + \frac{- 50}{8} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.31

Cancel the common factor of $- 50$ and $8$ .

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Step 15.4.31.1

Factor $2$ out of $- 50$ .

$\frac{- 9 + \frac{2 (- 25)}{8} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.31.2

Cancel the common factors.

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Step 15.4.31.2.1

Factor $2$ out of $8$ .

$\frac{- 9 + \frac{2 \cdot - 25}{2 \cdot 4} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.31.2.2

Cancel the common factor.

$\frac{- 9 + \frac{2 \cdot - 25}{2 \cdot 4} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.31.2.3

Rewrite the expression.

$\frac{- 9 + \frac{- 25}{4} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 15.4.32

Move the negative in front of the fraction.

$\frac{- 9 - \frac{25}{4} (\ln (| - 1 |) - \ln (| 3 |))}{2}$

Step 16

Simplify.

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Step 16.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{- 9 - \frac{25}{4} \ln (\frac{| - 1 |}{| 3 |})}{2}$

Step 16.2

Combine $\ln (\frac{| - 1 |}{| 3 |})$ and $\frac{25}{4}$ .

$\frac{- 9 - \frac{\ln (\frac{| - 1 |}{| 3 |}) \cdot 25}{4}}{2}$

Step 16.3

Move $25$ to the left of $\ln (\frac{| - 1 |}{| 3 |})$ .

$\frac{- 9 - \frac{25 \cdot \ln (\frac{| - 1 |}{| 3 |})}{4}}{2}$

Step 16.4

Rewrite $- 9$ as $- 1 (9)$ .

$\frac{- 1 (9) - \frac{25 \ln (\frac{| - 1 |}{| 3 |})}{4}}{2}$

Step 16.5

Factor $- 1$ out of $- \frac{25 \ln (\frac{| - 1 |}{| 3 |})}{4}$ .

$\frac{- 1 (9) - (\frac{25 \ln (\frac{| - 1 |}{| 3 |})}{4})}{2}$

Step 16.6

Factor $- 1$ out of $- 1 (9) - (\frac{25 \ln (\frac{| - 1 |}{| 3 |})}{4})$ .

$\frac{- 1 (9 + \frac{25 \ln (\frac{| - 1 |}{| 3 |})}{4})}{2}$

Step 16.7

Move the negative in front of the fraction.

$- \frac{9 + \frac{25 \ln (\frac{| - 1 |}{| 3 |})}{4}}{2}$

Step 17

Simplify.

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Step 17.1

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

$- \frac{9 + \frac{25 \ln (\frac{1}{| 3 |})}{4}}{2}$

Step 17.2

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$- \frac{9 + \frac{25 \ln (\frac{1}{3})}{4}}{2}$

Step 17.3

Simplify the numerator.

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Step 17.3.1

To write $9$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- \frac{9 \cdot \frac{4}{4} + \frac{25 \ln (\frac{1}{3})}{4}}{2}$

Step 17.3.2

Combine $9$ and $\frac{4}{4}$ .

$- \frac{\frac{9 \cdot 4}{4} + \frac{25 \ln (\frac{1}{3})}{4}}{2}$

Step 17.3.3

Combine the numerators over the common denominator.

$- \frac{\frac{9 \cdot 4 + 25 \ln (\frac{1}{3})}{4}}{2}$

Step 17.3.4

Multiply $9$ by $4$ .

$- \frac{\frac{36 + 25 \ln (\frac{1}{3})}{4}}{2}$

Step 17.4

Multiply the numerator by the reciprocal of the denominator.

$- (\frac{36 + 25 \ln (\frac{1}{3})}{4} \cdot \frac{1}{2})$

Step 17.5

Multiply $\frac{36 + 25 \ln (\frac{1}{3})}{4} \cdot \frac{1}{2}$ .

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Step 17.5.1

Multiply $\frac{36 + 25 \ln (\frac{1}{3})}{4}$ by $\frac{1}{2}$ .

$- \frac{36 + 25 \ln (\frac{1}{3})}{4 \cdot 2}$

Step 17.5.2

Multiply $4$ by $2$ .

$- \frac{36 + 25 \ln (\frac{1}{3})}{8}$

Step 18

The result can be shown in multiple forms.

Exact Form:

$- \frac{36 + 25 \ln (\frac{1}{3})}{8}$

Decimal Form:

$- 1.06683659 \dots$

Step 19