Calculus Examples

$\int_{0}^{1} \frac{30}{6 x^{2} + 7 x + 1} d x$

Step 1

Since $30$ is constant with respect to $x$ , move $30$ out of the integral.

$30 \int_{0}^{1} \frac{1}{6 x^{2} + 7 x + 1} d x$

Step 2

Write the fraction using partial fraction decomposition.

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Step 2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.1.1

Factor by grouping.

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Step 2.1.1.1

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 $a \cdot c = 6 \cdot 1 = 6$ and whose sum is $b = 7$ .

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Step 2.1.1.1.1

Factor $7$ out of $7 x$ .

$\frac{1}{6 x^{2} + 7 (x) + 1}$

Step 2.1.1.1.2

Rewrite $7$ as $1$ plus $6$

$\frac{1}{6 x^{2} + (1 + 6) x + 1}$

Step 2.1.1.1.3

Apply the distributive property.

$\frac{1}{6 x^{2} + 1 x + 6 x + 1}$

Step 2.1.1.1.4

Multiply $x$ by $1$ .

$\frac{1}{6 x^{2} + x + 6 x + 1}$

Step 2.1.1.2

Factor out the greatest common factor from each group.

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Step 2.1.1.2.1

Group the first two terms and the last two terms.

$\frac{1}{(6 x^{2} + x) + 6 x + 1}$

Step 2.1.1.2.2

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

$\frac{1}{x (6 x + 1) + 1 (6 x + 1)}$

Step 2.1.1.3

Factor the polynomial by factoring out the greatest common factor, $6 x + 1$ .

$\frac{1}{(6 x + 1) (x + 1)}$

Step 2.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{6 x + 1}$

Step 2.1.3

$\frac{A}{6 x + 1} + \frac{B}{x + 1}$

Step 2.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(6 x + 1) (x + 1)$ .

$\frac{1 (6 x + 1) (x + 1)}{(6 x + 1) (x + 1)} = \frac{(A) (6 x + 1) (x + 1)}{6 x + 1} + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.5

Cancel the common factor of $6 x + 1$ .

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Step 2.1.5.1

Cancel the common factor.

$\frac{1 (6 x + 1) (x + 1)}{(6 x + 1) (x + 1)} = \frac{(A) (6 x + 1) (x + 1)}{6 x + 1} + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.5.2

Rewrite the expression.

$\frac{1 (x + 1)}{x + 1} = \frac{(A) (6 x + 1) (x + 1)}{6 x + 1} + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.6

Cancel the common factor of $x + 1$ .

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Step 2.1.6.1

Cancel the common factor.

$\frac{1 (x + 1)}{x + 1} = \frac{(A) (6 x + 1) (x + 1)}{6 x + 1} + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.6.2

Rewrite the expression.

$1 = \frac{(A) (6 x + 1) (x + 1)}{6 x + 1} + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.7

Simplify each term.

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Step 2.1.7.1

Cancel the common factor of $6 x + 1$ .

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Step 2.1.7.1.1

Cancel the common factor.

$1 = \frac{A (6 x + 1) (x + 1)}{6 x + 1} + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.7.1.2

Divide $(A) (x + 1)$ by $1$ .

$1 = (A) (x + 1) + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.7.2

Apply the distributive property.

$1 = A x + A \cdot 1 + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.7.3

Multiply $A$ by $1$ .

$1 = A x + A + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.7.4

Cancel the common factor of $x + 1$ .

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Step 2.1.7.4.1

Cancel the common factor.

$1 = A x + A + \frac{(B) (6 x + 1) (x + 1)}{x + 1}$

Step 2.1.7.4.2

Divide $(B) (6 x + 1)$ by $1$ .

$1 = A x + A + (B) (6 x + 1)$

Step 2.1.7.5

Apply the distributive property.

$1 = A x + A + B (6 x) + B \cdot 1$

Step 2.1.7.6

Rewrite using the commutative property of multiplication.

$1 = A x + A + 6 B x + B \cdot 1$

Step 2.1.7.7

Multiply $B$ by $1$ .

$1 = A x + A + 6 B x + B$

Step 2.1.8

Simplify the expression.

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Step 2.1.8.1

Move $B$ .

$1 = A x + A + 6 x B + B$

Step 2.1.8.2

Move $A$ .

$1 = A x + 6 x B + A + B$

Step 2.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + 6 B$

Step 2.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A + B$

Step 2.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = A + 6 B$

$1 = A + B$

$0 = A + 6 B$

$1 = A + B$

Step 2.3

Solve the system of equations.

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Step 2.3.1

Solve for $A$ in $0 = A + 6 B$ .

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Step 2.3.1.1

Rewrite the equation as $A + 6 B = 0$ .

$A + 6 B = 0$

$1 = A + B$

Step 2.3.1.2

Subtract $6 B$ from both sides of the equation.

$A = - 6 B$

$1 = A + B$

$A = - 6 B$

$1 = A + B$

Step 2.3.2

Replace all occurrences of $A$ with $- 6 B$ in each equation.

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Step 2.3.2.1

Replace all occurrences of $A$ in $1 = A + B$ with $- 6 B$ .

$1 = (- 6 B) + B$

$A = - 6 B$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Add $- 6 B$ and $B$ .

$1 = - 5 B$

$A = - 6 B$

$1 = - 5 B$

$A = - 6 B$

$1 = - 5 B$

$A = - 6 B$

Step 2.3.3

Solve for $B$ in $1 = - 5 B$ .

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Step 2.3.3.1

Rewrite the equation as $- 5 B = 1$ .

$- 5 B = 1$

$A = - 6 B$

Step 2.3.3.2

Divide each term in $- 5 B = 1$ by $- 5$ and simplify.

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Step 2.3.3.2.1

Divide each term in $- 5 B = 1$ by $- 5$ .

$\frac{- 5 B}{- 5} = \frac{1}{- 5}$

$A = - 6 B$

Step 2.3.3.2.2

Simplify the left side.

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Step 2.3.3.2.2.1

Cancel the common factor of $- 5$ .

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Step 2.3.3.2.2.1.1

Cancel the common factor.

$\frac{- 5 B}{- 5} = \frac{1}{- 5}$

$A = - 6 B$

Step 2.3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{- 5}$

$A = - 6 B$

$B = \frac{1}{- 5}$

$A = - 6 B$

$B = \frac{1}{- 5}$

$A = - 6 B$

Step 2.3.3.2.3

Simplify the right side.

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Step 2.3.3.2.3.1

Move the negative in front of the fraction.

$B = - \frac{1}{5}$

$A = - 6 B$

$B = - \frac{1}{5}$

$A = - 6 B$

$B = - \frac{1}{5}$

$A = - 6 B$

$B = - \frac{1}{5}$

$A = - 6 B$

Step 2.3.4

Replace all occurrences of $B$ with $- \frac{1}{5}$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $B$ in $A = - 6 B$ with $- \frac{1}{5}$ .

$A = - 6 (- \frac{1}{5})$

$B = - \frac{1}{5}$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Multiply $- 6 (- \frac{1}{5})$ .

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Step 2.3.4.2.1.1

Multiply $- 1$ by $- 6$ .

$A = 6 (\frac{1}{5})$

$B = - \frac{1}{5}$

Step 2.3.4.2.1.2

Combine $6$ and $\frac{1}{5}$ .

$A = \frac{6}{5}$

$B = - \frac{1}{5}$

$A = \frac{6}{5}$

$B = - \frac{1}{5}$

$A = \frac{6}{5}$

$B = - \frac{1}{5}$

$A = \frac{6}{5}$

$B = - \frac{1}{5}$

Step 2.3.5

List all of the solutions.

$A = \frac{6}{5}, B = - \frac{1}{5}$

Step 2.4

Replace each of the partial fraction coefficients in $\frac{A}{6 x + 1} + \frac{B}{x + 1}$ with the values found for $A$ and $B$ .

$\frac{\frac{6}{5}}{6 x + 1} + \frac{- \frac{1}{5}}{x + 1}$

Step 2.5

Simplify.

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Step 2.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{6}{5} \cdot \frac{1}{6 x + 1} + \frac{- \frac{1}{5}}{x + 1}$

Step 2.5.2

Multiply $\frac{6}{5}$ by $\frac{1}{6 x + 1}$ .

$\frac{6}{5 (6 x + 1)} + \frac{- \frac{1}{5}}{x + 1}$

Step 2.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{6}{5 (6 x + 1)} - \frac{1}{5} \cdot \frac{1}{x + 1}$

Step 2.5.4

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

$\frac{6}{5 (6 x + 1)} - \frac{1}{(x + 1) \cdot 5}$

Step 2.5.5

Move $5$ to the left of $x + 1$ .

$30 \int_{0}^{1} \frac{6}{5 (6 x + 1)} - \frac{1}{5 (x + 1)} d x$

Step 3

Split the single integral into multiple integrals.

$30 (\int_{0}^{1} \frac{6}{5 (6 x + 1)} d x + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 4

Since $\frac{6}{5}$ is constant with respect to $x$ , move $\frac{6}{5}$ out of the integral.

$30 (\frac{6}{5} \int_{0}^{1} \frac{1}{6 x + 1} d x + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 5

Let $u_{1} = 6 x + 1$ . Then $d u_{1} = 6 d x$ , so $\frac{1}{6} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

Let $u_{1} = 6 x + 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $6 x + 1$ .

$\frac{d}{d x} [6 x + 1]$

Step 5.1.2

By the Sum Rule, the derivative of $6 x + 1$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [6 x] + \frac{d}{d x} [1]$

Step 5.1.3

Evaluate $\frac{d}{d x} [6 x]$ .

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Step 5.1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 5.1.3.2

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

$6 \cdot 1 + \frac{d}{d x} [1]$

Step 5.1.3.3

Multiply $6$ by $1$ .

$6 + \frac{d}{d x} [1]$

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$6 + 0$

Step 5.1.4.2

Add $6$ and $0$ .

$6$

Step 5.2

Substitute the lower limit in for $x$ in $u_{1} = 6 x + 1$ .

$u_{lower} = 6 \cdot 0 + 1$

Step 5.3

Simplify.

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Step 5.3.1

Multiply $6$ by $0$ .

$u_{lower} = 0 + 1$

Step 5.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 5.4

Substitute the upper limit in for $x$ in $u_{1} = 6 x + 1$ .

$u_{upper} = 6 \cdot 1 + 1$

Step 5.5

Simplify.

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Step 5.5.1

Multiply $6$ by $1$ .

$u_{upper} = 6 + 1$

Step 5.5.2

Add $6$ and $1$ .

$u_{upper} = 7$

Step 5.6

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

$u_{lower} = 1$

$u_{upper} = 7$

Step 5.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$30 (\frac{6}{5} \int_{1}^{7} \frac{1}{u_{1}} \cdot \frac{1}{6} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 6

Simplify.

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Step 6.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{6}$ .

$30 (\frac{6}{5} \int_{1}^{7} \frac{1}{u_{1} \cdot 6} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 6.2

Move $6$ to the left of $u_{1}$ .

$30 (\frac{6}{5} \int_{1}^{7} \frac{1}{6 u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 7

Since $\frac{1}{6}$ is constant with respect to $u_{1}$ , move $\frac{1}{6}$ out of the integral.

$30 (\frac{6}{5} (\frac{1}{6} \int_{1}^{7} \frac{1}{u_{1}} d u_{1}) + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 8

Simplify.

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Step 8.1

Multiply $\frac{1}{6}$ by $\frac{6}{5}$ .

$30 (\frac{6}{6 \cdot 5} \int_{1}^{7} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 8.2

Multiply $6$ by $5$ .

$30 (\frac{6}{30} \int_{1}^{7} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 8.3

Cancel the common factor of $6$ and $30$ .

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Step 8.3.1

Factor $6$ out of $6$ .

$30 (\frac{6 (1)}{30} \int_{1}^{7} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 8.3.2

Cancel the common factors.

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Step 8.3.2.1

Factor $6$ out of $30$ .

$30 (\frac{6 \cdot 1}{6 \cdot 5} \int_{1}^{7} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 8.3.2.2

Cancel the common factor.

$30 (\frac{6 \cdot 1}{6 \cdot 5} \int_{1}^{7} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 8.3.2.3

Rewrite the expression.

$30 (\frac{1}{5} \int_{1}^{7} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 9

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

$30 (\frac{1}{5} \ln (| u_{1} |)]_{1}^{7} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 10

Combine $\frac{1}{5}$ and $\ln (| u_{1} |)]_{1}^{7}$ .

$30 (\frac{\ln (| u_{1} |)]_{1}^{7}}{5} + \int_{0}^{1} - \frac{1}{5 (x + 1)} d x)$

Step 11

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

$30 (\frac{\ln (| u_{1} |)]_{1}^{7}}{5} - \int_{0}^{1} \frac{1}{5 (x + 1)} d x)$

Step 12

Since $\frac{1}{5}$ is constant with respect to $x$ , move $\frac{1}{5}$ out of the integral.

$30 (\frac{\ln (| u_{1} |)]_{1}^{7}}{5} - (\frac{1}{5} \int_{0}^{1} \frac{1}{x + 1} d x))$

Step 13

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 13.1

Let $u_{2} = x + 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 13.1.1

Differentiate $x + 1$ .

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

Step 13.1.2

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 13.1.3

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 13.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

Substitute the lower limit in for $x$ in $u_{2} = x + 1$ .

$u_{lower} = 0 + 1$

Step 13.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 13.4

Substitute the upper limit in for $x$ in $u_{2} = x + 1$ .

$u_{upper} = 1 + 1$

Step 13.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 13.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 13.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$30 (\frac{\ln (| u_{1} |)]_{1}^{7}}{5} - \frac{1}{5} \int_{1}^{2} \frac{1}{u_{2}} d u_{2})$

Step 14

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$30 (\frac{\ln (| u_{1} |)]_{1}^{7}}{5} - \frac{1}{5} \ln (| u_{2} |)]_{1}^{2})$

Step 15

Combine $\ln (| u_{2} |)]_{1}^{2}$ and $\frac{1}{5}$ .

$30 (\frac{\ln (| u_{1} |)]_{1}^{7}}{5} - \frac{\ln (| u_{2} |)]_{1}^{2}}{5})$

Step 16

Substitute and simplify.

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Step 16.1

Evaluate $\ln (| u_{1} |)$ at $7$ and at $1$ .

$30 (\frac{(\ln (| 7 |)) - \ln (| 1 |)}{5} - \frac{\ln (| u_{2} |)]_{1}^{2}}{5})$

Step 16.2

Evaluate $\ln (| u_{2} |)$ at $2$ and at $1$ .

$30 (\frac{(\ln (| 7 |)) - \ln (| 1 |)}{5} - \frac{(\ln (| 2 |)) - \ln (| 1 |)}{5})$

Step 16.3

Simplify.

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Step 16.3.1

Combine the numerators over the common denominator.

$30 \frac{\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |))}{5}$

Step 16.3.2

Combine $30$ and $\frac{\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |))}{5}$ .

$\frac{30 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |)))}{5}$

Step 16.3.3

Cancel the common factor of $30$ and $5$ .

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Step 16.3.3.1

Factor $5$ out of $30 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |)))$ .

$\frac{5 (6 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |))))}{5}$

Step 16.3.3.2

Cancel the common factors.

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Step 16.3.3.2.1

Factor $5$ out of $5$ .

$\frac{5 (6 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |))))}{5 (1)}$

Step 16.3.3.2.2

Cancel the common factor.

$\frac{5 (6 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |))))}{5 \cdot 1}$

Step 16.3.3.2.3

Rewrite the expression.

$\frac{6 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |)))}{1}$

Step 16.3.3.2.4

Divide $6 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |)))$ by $1$ .

$6 (\ln (| 7 |) - \ln (| 1 |) - (\ln (| 2 |) - \ln (| 1 |)))$

Step 17

Simplify.

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Step 17.1

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

$6 (\ln (\frac{| 7 |}{| 1 |}) - (\ln (| 2 |) - \ln (| 1 |)))$

Step 17.2

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

$6 (\ln (\frac{| 7 |}{| 1 |}) - \ln (\frac{| 2 |}{| 1 |}))$

Step 17.3

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

$6 \ln (\frac{\frac{| 7 |}{| 1 |}}{\frac{| 2 |}{| 1 |}})$

Step 17.4

Rewrite $\frac{\frac{| 7 |}{| 1 |}}{\frac{| 2 |}{| 1 |}}$ as a product.

$6 \ln (\frac{| 7 |}{| 1 |} \cdot \frac{1}{\frac{| 2 |}{| 1 |}})$

Step 17.5

Multiply by the reciprocal of the fraction to divide by $\frac{| 2 |}{| 1 |}$ .

$6 \ln (\frac{| 7 |}{| 1 |} (1 \frac{| 1 |}{| 2 |}))$

Step 17.6

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

$6 \ln (\frac{| 7 |}{| 1 |} \cdot \frac{| 1 |}{| 2 |})$

Step 17.7

Multiply $\frac{| 7 |}{| 1 |}$ by $\frac{| 1 |}{| 2 |}$ .

$6 \ln (\frac{| 7 | | 1 |}{| 1 | | 2 |})$

Step 17.8

To multiply absolute values, multiply the terms inside each absolute value.

$6 \ln (\frac{| 7 \cdot 1 |}{| 1 | | 2 |})$

Step 17.9

Multiply $7$ by $1$ .

$6 \ln (\frac{| 7 |}{| 1 | | 2 |})$

Step 17.10

To multiply absolute values, multiply the terms inside each absolute value.

$6 \ln (\frac{| 7 |}{| 1 \cdot 2 |})$

Step 17.11

Multiply $2$ by $1$ .

$6 \ln (\frac{| 7 |}{| 2 |})$

Step 18

Simplify.

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Step 18.1

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$6 \ln (\frac{7}{| 2 |})$

Step 18.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$6 \ln (\frac{7}{2})$

Step 19

The result can be shown in multiple forms.

Exact Form:

$6 \ln (\frac{7}{2})$

Decimal Form:

$7.51657781 \dots$

Step 20