Calculus Examples

$\int_{0}^{1} \frac{6 x + 2}{(x + 1) (2 x + 1)} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

Factor $2$ out of $6 x + 2$ .

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Step 1.1.1.1

Factor $2$ out of $6 x$ .

$\frac{2 (3 x) + 2}{(x + 1) (2 x + 1)}$

Step 1.1.1.2

Factor $2$ out of $2$ .

$\frac{2 (3 x) + 2 (1)}{(x + 1) (2 x + 1)}$

Step 1.1.1.3

Factor $2$ out of $2 (3 x) + 2 (1)$ .

$\frac{2 (3 x + 1)}{(x + 1) (2 x + 1)}$

Step 1.1.2

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 $A$ .

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

Step 1.1.3

$\frac{A}{x + 1} + \frac{B}{2 x + 1}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x + 1) (2 x + 1)$ .

$\frac{2 (3 x + 1) (x + 1) (2 x + 1)}{(x + 1) (2 x + 1)} = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.5

Cancel the common factor of $x + 1$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{2 (3 x + 1) (x + 1) (2 x + 1)}{(x + 1) (2 x + 1)} = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.5.2

Rewrite the expression.

$\frac{2 (3 x + 1) (2 x + 1)}{2 x + 1} = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.6

Cancel the common factor of $2 x + 1$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{2 (3 x + 1) (2 x + 1)}{2 x + 1} = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.6.2

Divide $2 (3 x + 1)$ by $1$ .

$2 (3 x + 1) = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.7

Apply the distributive property.

$2 (3 x) + 2 \cdot 1 = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.8

Multiply.

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Step 1.1.8.1

Multiply $3$ by $2$ .

$6 x + 2 \cdot 1 = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.8.2

Multiply $2$ by $1$ .

$6 x + 2 = \frac{(A) (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9

Simplify each term.

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Step 1.1.9.1

Cancel the common factor of $x + 1$ .

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Step 1.1.9.1.1

Cancel the common factor.

$6 x + 2 = \frac{A (x + 1) (2 x + 1)}{x + 1} + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9.1.2

Divide $(A) (2 x + 1)$ by $1$ .

$6 x + 2 = (A) (2 x + 1) + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9.2

Apply the distributive property.

$6 x + 2 = A (2 x) + A \cdot 1 + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9.3

Rewrite using the commutative property of multiplication.

$6 x + 2 = 2 A x + A \cdot 1 + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9.4

Multiply $A$ by $1$ .

$6 x + 2 = 2 A x + A + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9.5

Cancel the common factor of $2 x + 1$ .

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Step 1.1.9.5.1

Cancel the common factor.

$6 x + 2 = 2 A x + A + \frac{(B) (x + 1) (2 x + 1)}{2 x + 1}$

Step 1.1.9.5.2

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

$6 x + 2 = 2 A x + A + (B) (x + 1)$

Step 1.1.9.6

Apply the distributive property.

$6 x + 2 = 2 A x + A + B x + B \cdot 1$

Step 1.1.9.7

Multiply $B$ by $1$ .

$6 x + 2 = 2 A x + A + B x + B$

Step 1.1.10

Simplify the expression.

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Step 1.1.10.1

Move $A$ .

$6 x + 2 = 2 x A + A + B x + B$

Step 1.1.10.2

Reorder $B$ and $x$ .

$6 x + 2 = 2 x A + A + B x + B$

Step 1.1.10.3

Move $A$ .

$6 x + 2 = 2 x A + B x + A + B$

Step 1.2

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

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Step 1.2.1

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.

$6 = 2 A + B$

Step 1.2.2

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.

$2 = A + B$

Step 1.2.3

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

$6 = 2 A + B$

$2 = A + B$

$6 = 2 A + B$

$2 = A + B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $B$ in $6 = 2 A + B$ .

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Step 1.3.1.1

Rewrite the equation as $2 A + B = 6$ .

$2 A + B = 6$

$2 = A + B$

Step 1.3.1.2

Subtract $2 A$ from both sides of the equation.

$B = 6 - 2 A$

$2 = A + B$

$B = 6 - 2 A$

$2 = A + B$

Step 1.3.2

Replace all occurrences of $B$ with $6 - 2 A$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $B$ in $2 = A + B$ with $6 - 2 A$ .

$2 = A + 6 - 2 A$

$B = 6 - 2 A$

Step 1.3.2.2

Simplify $2 = A + 6 - 2 A$ .

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Step 1.3.2.2.1

Simplify the left side.

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Step 1.3.2.2.1.1

Remove parentheses.

$2 = A + 6 - 2 A$

$B = 6 - 2 A$

$2 = A + 6 - 2 A$

$B = 6 - 2 A$

Step 1.3.2.2.2

Simplify the right side.

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Step 1.3.2.2.2.1

Subtract $2 A$ from $A$ .

$2 = - A + 6$

$B = 6 - 2 A$

$2 = - A + 6$

$B = 6 - 2 A$

$2 = - A + 6$

$B = 6 - 2 A$

$2 = - A + 6$

$B = 6 - 2 A$

Step 1.3.3

Solve for $A$ in $2 = - A + 6$ .

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Step 1.3.3.1

Rewrite the equation as $- A + 6 = 2$ .

$- A + 6 = 2$

$B = 6 - 2 A$

Step 1.3.3.2

Move all terms not containing $A$ to the right side of the equation.

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Step 1.3.3.2.1

Subtract $6$ from both sides of the equation.

$- A = 2 - 6$

$B = 6 - 2 A$

Step 1.3.3.2.2

Subtract $6$ from $2$ .

$- A = - 4$

$B = 6 - 2 A$

$- A = - 4$

$B = 6 - 2 A$

Step 1.3.3.3

Divide each term in $- A = - 4$ by $- 1$ and simplify.

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Step 1.3.3.3.1

Divide each term in $- A = - 4$ by $- 1$ .

$\frac{- A}{- 1} = \frac{- 4}{- 1}$

$B = 6 - 2 A$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{- 4}{- 1}$

$B = 6 - 2 A$

Step 1.3.3.3.2.2

Divide $A$ by $1$ .

$A = \frac{- 4}{- 1}$

$B = 6 - 2 A$

$A = \frac{- 4}{- 1}$

$B = 6 - 2 A$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $- 4$ by $- 1$ .

$A = 4$

$B = 6 - 2 A$

$A = 4$

$B = 6 - 2 A$

$A = 4$

$B = 6 - 2 A$

$A = 4$

$B = 6 - 2 A$

Step 1.3.4

Replace all occurrences of $A$ with $4$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $A$ in $B = 6 - 2 A$ with $4$ .

$B = 6 - 2 \cdot 4$

$A = 4$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $6 - 2 \cdot 4$ .

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Step 1.3.4.2.1.1

Multiply $- 2$ by $4$ .

$B = 6 - 8$

$A = 4$

Step 1.3.4.2.1.2

Subtract $8$ from $6$ .

$B = - 2$

$A = 4$

$B = - 2$

$A = 4$

$B = - 2$

$A = 4$

$B = - 2$

$A = 4$

Step 1.3.5

List all of the solutions.

$B = - 2, A = 4$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x + 1} + \frac{B}{2 x + 1}$ with the values found for $A$ and $B$ .

$\frac{4}{x + 1} + \frac{- 2}{2 x + 1}$

Step 1.5

Move the negative in front of the fraction.

$\int_{0}^{1} \frac{4}{x + 1} - \frac{2}{2 x + 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{1} \frac{4}{x + 1} d x + \int_{0}^{1} - \frac{2}{2 x + 1} d x$

Step 3

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$4 \int_{0}^{1} \frac{1}{x + 1} d x + \int_{0}^{1} - \frac{2}{2 x + 1} d x$

Step 4

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

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Step 4.1

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

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Step 4.1.1

Differentiate $x + 1$ .

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

Step 4.1.2

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 4.1.3

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

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

Step 4.1.4

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

$1 + 0$

Step 4.1.5

Add $1$ and $0$ .

$1$

Step 4.2

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

$u_{lower} = 0 + 1$

Step 4.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 4.4

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

$u_{upper} = 1 + 1$

Step 4.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 4.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 4.7

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

$4 \int_{1}^{2} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} - \frac{2}{2 x + 1} d x$

Step 5

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

$4 (\ln (| u_{1} |)]_{1}^{2}) + \int_{0}^{1} - \frac{2}{2 x + 1} d x$

Step 6

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

$4 (\ln (| u_{1} |)]_{1}^{2}) - \int_{0}^{1} \frac{2}{2 x + 1} d x$

Step 7

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$4 (\ln (| u_{1} |)]_{1}^{2}) - (2 \int_{0}^{1} \frac{1}{2 x + 1} d x)$

Step 8

Multiply $2$ by $- 1$ .

$4 (\ln (| u_{1} |)]_{1}^{2}) - 2 \int_{0}^{1} \frac{1}{2 x + 1} d x$

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $2 x + 1$ .

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

Step 9.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 9.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 9.1.3.1

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

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Multiply $2$ by $1$ .

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

Step 9.1.4

Differentiate using the Constant Rule.

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Step 9.1.4.1

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

$2 + 0$

Step 9.1.4.2

Add $2$ and $0$ .

$2$

Step 9.2

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

$u_{lower} = 2 \cdot 0 + 1$

Step 9.3

Simplify.

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Step 9.3.1

Multiply $2$ by $0$ .

$u_{lower} = 0 + 1$

Step 9.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 9.4

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

$u_{upper} = 2 \cdot 1 + 1$

Step 9.5

Simplify.

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Step 9.5.1

Multiply $2$ by $1$ .

$u_{upper} = 2 + 1$

Step 9.5.2

Add $2$ and $1$ .

$u_{upper} = 3$

Step 9.6

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

$u_{lower} = 1$

$u_{upper} = 3$

Step 9.7

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

$4 (\ln (| u_{1} |)]_{1}^{2}) - 2 \int_{1}^{3} \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 10

Simplify.

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Step 10.1

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

$4 (\ln (| u_{1} |)]_{1}^{2}) - 2 \int_{1}^{3} \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 10.2

Move $2$ to the left of $u_{2}$ .

$4 (\ln (| u_{1} |)]_{1}^{2}) - 2 \int_{1}^{3} \frac{1}{2 u_{2}} d u_{2}$

Step 11

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

$4 (\ln (| u_{1} |)]_{1}^{2}) - 2 (\frac{1}{2} \int_{1}^{3} \frac{1}{u_{2}} d u_{2})$

Step 12

Simplify.

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Step 12.1

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

$4 (\ln (| u_{1} |)]_{1}^{2}) + \frac{- 2}{2} \int_{1}^{3} \frac{1}{u_{2}} d u_{2}$

Step 12.2

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

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Step 12.2.1

Factor $2$ out of $- 2$ .

$4 (\ln (| u_{1} |)]_{1}^{2}) + \frac{2 \cdot - 1}{2} \int_{1}^{3} \frac{1}{u_{2}} d u_{2}$

Step 12.2.2

Cancel the common factors.

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Step 12.2.2.1

Factor $2$ out of $2$ .

$4 (\ln (| u_{1} |)]_{1}^{2}) + \frac{2 \cdot - 1}{2 (1)} \int_{1}^{3} \frac{1}{u_{2}} d u_{2}$

Step 12.2.2.2

Cancel the common factor.

$4 (\ln (| u_{1} |)]_{1}^{2}) + \frac{2 \cdot - 1}{2 \cdot 1} \int_{1}^{3} \frac{1}{u_{2}} d u_{2}$

Step 12.2.2.3

Rewrite the expression.

$4 (\ln (| u_{1} |)]_{1}^{2}) + \frac{- 1}{1} \int_{1}^{3} \frac{1}{u_{2}} d u_{2}$

Step 12.2.2.4

Divide $- 1$ by $1$ .

$4 (\ln (| u_{1} |)]_{1}^{2}) - \int_{1}^{3} \frac{1}{u_{2}} d u_{2}$

Step 13

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

$4 (\ln (| u_{1} |)]_{1}^{2}) - (\ln (| u_{2} |)]_{1}^{3})$

Step 14

Substitute and simplify.

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Step 14.1

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

$4 ((\ln (| 2 |)) - \ln (| 1 |)) - (\ln (| u_{2} |)]_{1}^{3})$

Step 14.2

Evaluate $\ln (| u_{2} |)$ at $3$ and at $1$ .

$4 ((\ln (| 2 |)) - \ln (| 1 |)) - ((\ln (| 3 |)) - \ln (| 1 |))$

Step 14.3

Remove parentheses.

$4 (\ln (| 2 |) - \ln (| 1 |)) - (\ln (| 3 |) - \ln (| 1 |))$

Step 15

Simplify.

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Step 15.1

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

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

Step 15.2

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

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

Step 16

Simplify.

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Step 16.1

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

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

Step 16.2

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

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

Step 16.3

Divide $2$ by $1$ .

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

Step 16.4

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

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

Step 16.5

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

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

Step 16.6

Divide $3$ by $1$ .

$4 \ln (2) - \ln (3)$

Step 17

The result can be shown in multiple forms.

Exact Form:

$4 \ln (2) - \ln (3)$

Decimal Form:

$1.67397643 \dots$

Step 18