Calculus Examples

$\int_{4}^{5} \frac{3 x^{2}}{{(x - 2)}^{2} (x - 3)} d x$

Step 1

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$3 \int_{4}^{5} \frac{x^{2}}{{(x - 2)}^{2} (x - 3)} d x$

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

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 $A$ .

$\frac{A}{{(x - 2)}^{2}}$

Step 2.1.2

$\frac{A}{{(x - 2)}^{2}} + \frac{B}{x - 2}$

Step 2.1.3

$\frac{A}{{(x - 2)}^{2}} + \frac{B}{x - 2} + \frac{C}{x - 3}$

Step 2.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x - 2)}^{2} (x - 3)$ .

$\frac{x^{2} {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2} (x - 3)} = \frac{(A) {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.5

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

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Step 2.1.5.1

Cancel the common factor.

$\frac{x^{2} {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2} (x - 3)} = \frac{(A) {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.5.2

Rewrite the expression.

$\frac{x^{2} (x - 3)}{x - 3} = \frac{(A) {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.6

Cancel the common factor of $x - 3$ .

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Step 2.1.6.1

Cancel the common factor.

$\frac{x^{2} (x - 3)}{x - 3} = \frac{(A) {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.6.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{(A) {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7

Simplify each term.

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Step 2.1.7.1

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

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Step 2.1.7.1.1

Cancel the common factor.

$x^{2} = \frac{A {(x - 2)}^{2} (x - 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.1.2

Divide $(A) (x - 3)$ by $1$ .

$x^{2} = (A) (x - 3) + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.2

Apply the distributive property.

$x^{2} = A x + A \cdot - 3 + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.3

Move $- 3$ to the left of $A$ .

$x^{2} = A x - 3 \cdot A + \frac{(B) {(x - 2)}^{2} (x - 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.4

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

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Step 2.1.7.4.1

Factor $x - 2$ out of $(B) {(x - 2)}^{2} (x - 3)$ .

$x^{2} = A x - 3 A + \frac{(x - 2) (B (x - 2) (x - 3))}{x - 2} + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.4.2

Cancel the common factors.

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Step 2.1.7.4.2.1

Multiply by $1$ .

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

Step 2.1.7.4.2.2

Cancel the common factor.

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

Step 2.1.7.4.2.3

Rewrite the expression.

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

Step 2.1.7.4.2.4

Divide $B (x - 2) (x - 3)$ by $1$ .

$x^{2} = A x - 3 A + B (x - 2) (x - 3) + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.5

Apply the distributive property.

$x^{2} = A x - 3 A + (B x + B \cdot - 2) (x - 3) + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.6

Move $- 2$ to the left of $B$ .

$x^{2} = A x - 3 A + (B x - 2 \cdot B) (x - 3) + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.7

Expand $(B x - 2 B) (x - 3)$ using the FOIL Method.

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Step 2.1.7.7.1

Apply the distributive property.

$x^{2} = A x - 3 A + B x (x - 3) - 2 B (x - 3) + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.7.2

Apply the distributive property.

$x^{2} = A x - 3 A + B x \cdot x + B x \cdot - 3 - 2 B (x - 3) + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.7.3

Apply the distributive property.

$x^{2} = A x - 3 A + B x \cdot x + B x \cdot - 3 - 2 B x - 2 B \cdot - 3 + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.8

Simplify and combine like terms.

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Step 2.1.7.8.1

Simplify each term.

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Step 2.1.7.8.1.1

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

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Step 2.1.7.8.1.1.1

Move $x$ .

$x^{2} = A x - 3 A + B (x \cdot x) + B x \cdot - 3 - 2 B x - 2 B \cdot - 3 + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.8.1.1.2

Multiply $x$ by $x$ .

$x^{2} = A x - 3 A + B x^{2} + B x \cdot - 3 - 2 B x - 2 B \cdot - 3 + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.8.1.2

Move $- 3$ to the left of $B x$ .

$x^{2} = A x - 3 A + B x^{2} - 3 \cdot (B x) - 2 B x - 2 B \cdot - 3 + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.8.1.3

Multiply $- 3$ by $- 2$ .

$x^{2} = A x - 3 A + B x^{2} - 3 B x - 2 B x + 6 B + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.8.2

Subtract $2 B x$ from $- 3 B x$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.9

Cancel the common factor of $x - 3$ .

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Step 2.1.7.9.1

Cancel the common factor.

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + \frac{(C) {(x - 2)}^{2} (x - 3)}{x - 3}$

Step 2.1.7.9.2

Divide $(C) {(x - 2)}^{2}$ by $1$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + (C) {(x - 2)}^{2}$

Step 2.1.7.10

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C ((x - 2) (x - 2))$

Step 2.1.7.11

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

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Step 2.1.7.11.1

Apply the distributive property.

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x (x - 2) - 2 (x - 2))$

Step 2.1.7.11.2

Apply the distributive property.

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x \cdot x + x \cdot - 2 - 2 (x - 2))$

Step 2.1.7.11.3

Apply the distributive property.

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 2.1.7.12

Simplify and combine like terms.

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Step 2.1.7.12.1

Simplify each term.

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Step 2.1.7.12.1.1

Multiply $x$ by $x$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 2.1.7.12.1.2

Move $- 2$ to the left of $x$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2)$

Step 2.1.7.12.1.3

Multiply $- 2$ by $- 2$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x^{2} - 2 x - 2 x + 4)$

Step 2.1.7.12.2

Subtract $2 x$ from $- 2 x$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C (x^{2} - 4 x + 4)$

Step 2.1.7.13

Apply the distributive property.

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C x^{2} + C (- 4 x) + C \cdot 4$

Step 2.1.7.14

Simplify.

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Step 2.1.7.14.1

Rewrite using the commutative property of multiplication.

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C x^{2} - 4 C x + C \cdot 4$

Step 2.1.7.14.2

Move $4$ to the left of $C$ .

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C x^{2} - 4 C x + 4 \cdot C$

$x^{2} = A x - 3 A + B x^{2} - 5 B x + 6 B + C x^{2} - 4 C x + 4 C$

Step 2.1.8

Simplify the expression.

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Step 2.1.8.1

Move $B$ .

$x^{2} = A x - 3 A + B x^{2} - 5 x B + 6 B + C x^{2} - 4 C x + 4 C$

Step 2.1.8.2

Move $6 B$ .

$x^{2} = A x - 3 A + B x^{2} - 5 x B + C x^{2} - 4 C x + 6 B + 4 C$

Step 2.1.8.3

Move $- 3 A$ .

$x^{2} = A x + B x^{2} - 5 x B + C x^{2} - 4 C x - 3 A + 6 B + 4 C$

Step 2.1.8.4

Move $- 5 x B$ .

$x^{2} = A x + B x^{2} + C x^{2} - 5 x B - 4 C x - 3 A + 6 B + 4 C$

Step 2.1.8.5

Move $A x$ .

$x^{2} = B x^{2} + C x^{2} + A x - 5 x B - 4 C x - 3 A + 6 B + 4 C$

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = B + C$

Step 2.2.2

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.

$0 = A - 5 B - 4 C$

Step 2.2.3

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.

$0 = - 3 A + 6 B + 4 C$

Step 2.2.4

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

$1 = B + C$

$0 = A - 5 B - 4 C$

$0 = - 3 A + 6 B + 4 C$

$1 = B + C$

$0 = A - 5 B - 4 C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3

Solve the system of equations.

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Step 2.3.1

Solve for $B$ in $1 = B + C$ .

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Step 2.3.1.1

Rewrite the equation as $B + C = 1$ .

$B + C = 1$

$0 = A - 5 B - 4 C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.1.2

Subtract $C$ from both sides of the equation.

$B = 1 - C$

$0 = A - 5 B - 4 C$

$0 = - 3 A + 6 B + 4 C$

$B = 1 - C$

$0 = A - 5 B - 4 C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.2

Replace all occurrences of $B$ with $1 - C$ in each equation.

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Step 2.3.2.1

Replace all occurrences of $B$ in $0 = A - 5 B - 4 C$ with $1 - C$ .

$0 = A - 5 (1 - C) - 4 C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Simplify $A - 5 (1 - C) - 4 C$ .

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Step 2.3.2.2.1.1

Simplify each term.

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Step 2.3.2.2.1.1.1

Apply the distributive property.

$0 = A - 5 \cdot 1 - 5 (- C) - 4 C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.2.2.1.1.2

Multiply $- 5$ by $1$ .

$0 = A - 5 - 5 (- C) - 4 C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.2.2.1.1.3

Multiply $- 1$ by $- 5$ .

$0 = A - 5 + 5 C - 4 C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

$0 = A - 5 + 5 C - 4 C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.2.2.1.2

Subtract $4 C$ from $5 C$ .

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 B + 4 C$

Step 2.3.2.3

Replace all occurrences of $B$ in $0 = - 3 A + 6 B + 4 C$ with $1 - C$ .

$0 = - 3 A + 6 (1 - C) + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

Step 2.3.2.4

Simplify the right side.

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Step 2.3.2.4.1

Simplify $- 3 A + 6 (1 - C) + 4 C$ .

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Step 2.3.2.4.1.1

Simplify each term.

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Step 2.3.2.4.1.1.1

Apply the distributive property.

$0 = - 3 A + 6 \cdot 1 + 6 (- C) + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

Step 2.3.2.4.1.1.2

Multiply $6$ by $1$ .

$0 = - 3 A + 6 + 6 (- C) + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

Step 2.3.2.4.1.1.3

Multiply $- 1$ by $6$ .

$0 = - 3 A + 6 - 6 C + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 - 6 C + 4 C$

$0 = A - 5 + C$

$B = 1 - C$

Step 2.3.2.4.1.2

Add $- 6 C$ and $4 C$ .

$0 = - 3 A + 6 - 2 C$

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 - 2 C$

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 - 2 C$

$0 = A - 5 + C$

$B = 1 - C$

$0 = - 3 A + 6 - 2 C$

$0 = A - 5 + C$

$B = 1 - C$

Step 2.3.3

Reorder $1$ and $- C$ .

$B = - C + 1$

$0 = - 3 A + 6 - 2 C$

$0 = A - 5 + C$

Step 2.3.4

Solve for $A$ in $0 = A - 5 + C$ .

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Step 2.3.4.1

Rewrite the equation as $A - 5 + C = 0$ .

$A - 5 + C = 0$

$B = - C + 1$

$0 = - 3 A + 6 - 2 C$

Step 2.3.4.2

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

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Step 2.3.4.2.1

Add $5$ to both sides of the equation.

$A + C = 5$

$B = - C + 1$

$0 = - 3 A + 6 - 2 C$

Step 2.3.4.2.2

Subtract $C$ from both sides of the equation.

$A = 5 - C$

$B = - C + 1$

$0 = - 3 A + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

$0 = - 3 A + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

$0 = - 3 A + 6 - 2 C$

Step 2.3.5

Replace all occurrences of $A$ with $5 - C$ in each equation.

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Step 2.3.5.1

Replace all occurrences of $A$ in $0 = - 3 A + 6 - 2 C$ with $5 - C$ .

$0 = - 3 (5 - C) + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

Step 2.3.5.2

Simplify the right side.

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Step 2.3.5.2.1

Simplify $- 3 (5 - C) + 6 - 2 C$ .

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Step 2.3.5.2.1.1

Simplify each term.

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Step 2.3.5.2.1.1.1

Apply the distributive property.

$0 = - 3 \cdot 5 - 3 (- C) + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

Step 2.3.5.2.1.1.2

Multiply $- 3$ by $5$ .

$0 = - 15 - 3 (- C) + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

Step 2.3.5.2.1.1.3

Multiply $- 1$ by $- 3$ .

$0 = - 15 + 3 C + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

$0 = - 15 + 3 C + 6 - 2 C$

$A = 5 - C$

$B = - C + 1$

Step 2.3.5.2.1.2

Simplify by adding terms.

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Step 2.3.5.2.1.2.1

Add $- 15$ and $6$ .

$0 = 3 C - 9 - 2 C$

$A = 5 - C$

$B = - C + 1$

Step 2.3.5.2.1.2.2

Subtract $2 C$ from $3 C$ .

$0 = C - 9$

$A = 5 - C$

$B = - C + 1$

$0 = C - 9$

$A = 5 - C$

$B = - C + 1$

$0 = C - 9$

$A = 5 - C$

$B = - C + 1$

$0 = C - 9$

$A = 5 - C$

$B = - C + 1$

$0 = C - 9$

$A = 5 - C$

$B = - C + 1$

Step 2.3.6

Solve for $C$ in $0 = C - 9$ .

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Step 2.3.6.1

Rewrite the equation as $C - 9 = 0$ .

$C - 9 = 0$

$A = 5 - C$

$B = - C + 1$

Step 2.3.6.2

Add $9$ to both sides of the equation.

$C = 9$

$A = 5 - C$

$B = - C + 1$

$C = 9$

$A = 5 - C$

$B = - C + 1$

Step 2.3.7

Replace all occurrences of $C$ with $9$ in each equation.

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Step 2.3.7.1

Replace all occurrences of $C$ in $A = 5 - C$ with $9$ .

$A = 5 - (9)$

$C = 9$

$B = - C + 1$

Step 2.3.7.2

Simplify the right side.

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Step 2.3.7.2.1

Simplify $5 - (9)$ .

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Step 2.3.7.2.1.1

Multiply $- 1$ by $9$ .

$A = 5 - 9$

$C = 9$

$B = - C + 1$

Step 2.3.7.2.1.2

Subtract $9$ from $5$ .

$A = - 4$

$C = 9$

$B = - C + 1$

$A = - 4$

$C = 9$

$B = - C + 1$

$A = - 4$

$C = 9$

$B = - C + 1$

Step 2.3.7.3

Replace all occurrences of $C$ in $B = - C + 1$ with $9$ .

$B = - (9) + 1$

$A = - 4$

$C = 9$

Step 2.3.7.4

Simplify the right side.

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Step 2.3.7.4.1

Simplify $- (9) + 1$ .

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Step 2.3.7.4.1.1

Multiply $- 1$ by $9$ .

$B = - 9 + 1$

$A = - 4$

$C = 9$

Step 2.3.7.4.1.2

Add $- 9$ and $1$ .

$B = - 8$

$A = - 4$

$C = 9$

$B = - 8$

$A = - 4$

$C = 9$

$B = - 8$

$A = - 4$

$C = 9$

$B = - 8$

$A = - 4$

$C = 9$

Step 2.3.8

List all of the solutions.

$B = - 8, A = - 4, C = 9$

Step 2.4

Replace each of the partial fraction coefficients in $\frac{A}{{(x - 2)}^{2}} + \frac{B}{x - 2} + \frac{C}{x - 3}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{- 4}{{(x - 2)}^{2}} + \frac{- 8}{x - 2} + \frac{9}{x - 3}$

Step 2.5

Simplify.

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Step 2.5.1

Move the negative in front of the fraction.

$- \frac{4}{{(x - 2)}^{2}} + \frac{- 8}{x - 2} + \frac{9}{x - 3}$

Step 2.5.2

Move the negative in front of the fraction.

$3 \int_{4}^{5} - \frac{4}{{(x - 2)}^{2}} - \frac{8}{x - 2} + \frac{9}{x - 3} d x$

Step 3

Split the single integral into multiple integrals.

$3 (\int_{4}^{5} - \frac{4}{{(x - 2)}^{2}} d x + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 4

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

$3 (- \int_{4}^{5} \frac{4}{{(x - 2)}^{2}} d x + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 5

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

$3 (- (4 \int_{4}^{5} \frac{1}{{(x - 2)}^{2}} d x) + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 6

Multiply $4$ by $- 1$ .

$3 (- 4 \int_{4}^{5} \frac{1}{{(x - 2)}^{2}} d x + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x - 2$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$u_{lower} = 4 - 2$

Step 7.3

Subtract $2$ from $4$ .

$u_{lower} = 2$

Step 7.4

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

$u_{upper} = 5 - 2$

Step 7.5

Subtract $2$ from $5$ .

$u_{upper} = 3$

Step 7.6

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

$u_{lower} = 2$

$u_{upper} = 3$

Step 7.7

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

$3 (- 4 \int_{2}^{3} \frac{1}{{u_{1}}^{2}} d u_{1} + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 8

Apply basic rules of exponents.

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Step 8.1

Move ${u_{1}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$3 (- 4 \int_{2}^{3} {({u_{1}}^{2})}^{- 1} d u_{1} + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 8.2

Multiply the exponents in ${({u_{1}}^{2})}^{- 1}$ .

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Step 8.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$3 (- 4 \int_{2}^{3} {u_{1}}^{2 \cdot - 1} d u_{1} + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 8.2.2

Multiply $2$ by $- 1$ .

$3 (- 4 \int_{2}^{3} {u_{1}}^{- 2} d u_{1} + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 9

By the Power Rule, the integral of ${u_{1}}^{- 2}$ with respect to $u_{1}$ is $- {u_{1}}^{- 1}$ .

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) + \int_{4}^{5} - \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 10

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

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - \int_{4}^{5} \frac{8}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 11

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

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - (8 \int_{4}^{5} \frac{1}{x - 2} d x) + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 12

Multiply $8$ by $- 1$ .

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - 8 \int_{4}^{5} \frac{1}{x - 2} d x + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Differentiate $x - 2$ .

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

Step 13.1.2

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 13.1.3

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

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

Step 13.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ 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 $u_{2} = x - 2$ .

$u_{lower} = 4 - 2$

Step 13.3

Subtract $2$ from $4$ .

$u_{lower} = 2$

Step 13.4

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

$u_{upper} = 5 - 2$

Step 13.5

Subtract $2$ from $5$ .

$u_{upper} = 3$

Step 13.6

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

$u_{lower} = 2$

$u_{upper} = 3$

Step 13.7

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

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - 8 \int_{2}^{3} \frac{1}{u_{2}} d u_{2} + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 14

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

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - 8 (\ln (| u_{2} |)]_{2}^{3}) + \int_{4}^{5} \frac{9}{x - 3} d x)$

Step 15

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - 8 (\ln (| u_{2} |)]_{2}^{3}) + 9 \int_{4}^{5} \frac{1}{x - 3} d x)$

Step 16

Let $u_{3} = x - 3$ . Then $d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 16.1

Let $u_{3} = x - 3$ . Find $\frac{d u_{3}}{d x}$ .

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Step 16.1.1

Differentiate $x - 3$ .

$\frac{d}{d x} [x - 3]$

Step 16.1.2

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 16.1.3

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

$1 + \frac{d}{d x} [- 3]$

Step 16.1.4

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$1 + 0$

Step 16.1.5

Add $1$ and $0$ .

$1$

Step 16.2

Substitute the lower limit in for $x$ in $u_{3} = x - 3$ .

$u_{lower} = 4 - 3$

Step 16.3

Subtract $3$ from $4$ .

$u_{lower} = 1$

Step 16.4

Substitute the upper limit in for $x$ in $u_{3} = x - 3$ .

$u_{upper} = 5 - 3$

Step 16.5

Subtract $3$ from $5$ .

$u_{upper} = 2$

Step 16.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 16.7

Rewrite the problem using $u_{3}$ , $d u_{3}$ , and the new limits of integration.

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - 8 (\ln (| u_{2} |)]_{2}^{3}) + 9 \int_{1}^{2} \frac{1}{u_{3}} d u_{3})$

Step 17

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$3 (- 4 (- {u_{1}}^{- 1}]_{2}^{3}) - 8 (\ln (| u_{2} |)]_{2}^{3}) + 9 (\ln (| u_{3} |)]_{1}^{2}))$

Step 18

Substitute and simplify.

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Step 18.1

Evaluate $- {u_{1}}^{- 1}$ at $3$ and at $2$ .

$3 (- 4 ((- 3^{- 1}) + 2^{- 1}) - 8 (\ln (| u_{2} |)]_{2}^{3}) + 9 (\ln (| u_{3} |)]_{1}^{2}))$

Step 18.2

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

$3 (- 4 ((- 3^{- 1}) + 2^{- 1}) - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 (\ln (| u_{3} |)]_{1}^{2}))$

Step 18.3

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

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

Step 18.4

Simplify.

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Step 18.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 18.4.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 18.4.3

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 18.4.4

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 18.4.5

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 18.4.5.1

Multiply $\frac{1}{3}$ by $\frac{2}{2}$ .

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

Step 18.4.5.2

Multiply $3$ by $2$ .

$3 (- 4 (- \frac{2}{6} + \frac{1}{2} \cdot \frac{3}{3}) - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.5.3

Multiply $\frac{1}{2}$ by $\frac{3}{3}$ .

$3 (- 4 (- \frac{2}{6} + \frac{3}{2 \cdot 3}) - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.5.4

Multiply $2$ by $3$ .

$3 (- 4 (- \frac{2}{6} + \frac{3}{6}) - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.6

Combine the numerators over the common denominator.

$3 (- 4 \frac{- 2 + 3}{6} - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.7

Add $- 2$ and $3$ .

$3 (- 4 (\frac{1}{6}) - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.8

Combine $- 4$ and $\frac{1}{6}$ .

$3 (\frac{- 4}{6} - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.9

Cancel the common factor of $- 4$ and $6$ .

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Step 18.4.9.1

Factor $2$ out of $- 4$ .

$3 (\frac{2 (- 2)}{6} - 8 ((\ln (| 3 |)) - \ln (| 2 |)) + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.9.2

Cancel the common factors.

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Step 18.4.9.2.1

Factor $2$ out of $6$ .

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

Step 18.4.9.2.2

Cancel the common factor.

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

Step 18.4.9.2.3

Rewrite the expression.

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

Step 18.4.10

Move the negative in front of the fraction.

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

Step 18.4.11

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

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

Step 18.4.12

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

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

Step 18.4.13

Combine the numerators over the common denominator.

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

Step 18.4.14

Multiply $3$ by $- 8$ .

$3 (\frac{- 2 - 24 (\ln (| 3 |) - \ln (| 2 |))}{3} + 9 ((\ln (| 2 |)) - \ln (| 1 |)))$

Step 18.4.15

To write $9 (\ln (| 2 |) - \ln (| 1 |))$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$3 (\frac{- 2 - 24 (\ln (| 3 |) - \ln (| 2 |))}{3} + 9 (\ln (| 2 |) - \ln (| 1 |)) \cdot \frac{3}{3})$

Step 18.4.16

Combine $9 (\ln (| 2 |) - \ln (| 1 |))$ and $\frac{3}{3}$ .

$3 (\frac{- 2 - 24 (\ln (| 3 |) - \ln (| 2 |))}{3} + \frac{9 (\ln (| 2 |) - \ln (| 1 |)) \cdot 3}{3})$

Step 18.4.17

Combine the numerators over the common denominator.

$3 \frac{- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 9 (\ln (| 2 |) - \ln (| 1 |)) \cdot 3}{3}$

Step 18.4.18

Multiply $3$ by $9$ .

$3 \frac{- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 27 (\ln (| 2 |) - \ln (| 1 |))}{3}$

Step 18.4.19

Combine $3$ and $\frac{- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 27 (\ln (| 2 |) - \ln (| 1 |))}{3}$ .

$\frac{3 (- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 27 (\ln (| 2 |) - \ln (| 1 |)))}{3}$

Step 18.4.20

Cancel the common factor of $3$ .

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Step 18.4.20.1

Cancel the common factor.

$\frac{3 (- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 27 (\ln (| 2 |) - \ln (| 1 |)))}{3}$

Step 18.4.20.2

Divide $- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 27 (\ln (| 2 |) - \ln (| 1 |))$ by $1$ .

$- 2 - 24 (\ln (| 3 |) - \ln (| 2 |)) + 27 (\ln (| 2 |) - \ln (| 1 |))$

Step 19

Simplify.

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Step 19.1

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

$- 2 - 24 \ln (\frac{| 3 |}{| 2 |}) + 27 (\ln (| 2 |) - \ln (| 1 |))$

Step 19.2

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

$- 2 - 24 \ln (\frac{| 3 |}{| 2 |}) + 27 \ln (\frac{| 2 |}{| 1 |})$

Step 20

Simplify.

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Step 20.1

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

$- 2 - 24 \ln (\frac{3}{| 2 |}) + 27 \ln (\frac{| 2 |}{| 1 |})$

Step 20.2

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

$- 2 - 24 \ln (\frac{3}{2}) + 27 \ln (\frac{| 2 |}{| 1 |})$

Step 20.3

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

$- 2 - 24 \ln (\frac{3}{2}) + 27 \ln (\frac{2}{| 1 |})$

Step 20.4

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

$- 2 - 24 \ln (\frac{3}{2}) + 27 \ln (\frac{2}{1})$

Step 20.5

Divide $2$ by $1$ .

$- 2 - 24 \ln (\frac{3}{2}) + 27 \ln (2)$

Step 21

The result can be shown in multiple forms.

Exact Form:

$- 2 - 24 \ln (\frac{3}{2}) + 27 \ln (2)$

Decimal Form:

$6.98381128 \dots$

Step 22