Calculus Examples

$\int_{4}^{5} \frac{x^{3} - 3 x^{2} - 9}{x^{3} - 3 x^{2}} d x$

Step 1

Divide $x^{3} - 3 x^{2} - 9$ by $x^{3} - 3 x^{2}$ .

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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 $0$ .

$x^{3}$

$3 x^{2}$

$0 x$

$0$

$x^{3}$

$3 x^{2}$

$0 x$

$9$

Step 1.2

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x^{3}$ .

									$1$
	$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$0$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	-	$9$

Step 1.3

Multiply the new quotient term by the divisor.

$1$

$x^{3}$

$3 x^{2}$

$0 x$

$0$

$x^{3}$

$3 x^{2}$

$0 x$

$9$

$x^{3}$

$3 x^{2}$

$0$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 3 x^{2} + 0 + 0$

								$1$
$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$0$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	-	$9$
							-	$x^{3}$	+	$3 x^{2}$	-	$0$	-	$0$

Step 1.5

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

								$1$
$x^{3}$	-	$3 x^{2}$	+	$0 x$	+	$0$		$x^{3}$	-	$3 x^{2}$	+	$0 x$	-	$9$
							-	$x^{3}$	+	$3 x^{2}$	-	$0$	-	$0$
													-	$9$

Step 1.6

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

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

Apply the constant rule.

$x]_{4}^{5} + \int_{4}^{5} - \frac{9}{x^{3} - 3 x^{2}} d x$

Step 4

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

$x]_{4}^{5} - \int_{4}^{5} \frac{9}{x^{3} - 3 x^{2}} d x$

Step 5

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

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

Step 6

Multiply $9$ by $- 1$ .

$x]_{4}^{5} - 9 \int_{4}^{5} \frac{1}{x^{3} - 3 x^{2}} d x$

Step 7

Write the fraction using partial fraction decomposition.

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Step 7.1

Decompose the fraction and multiply through by the common denominator.

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Step 7.1.1

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

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Step 7.1.1.1

Factor $x^{2}$ out of $x^{3}$ .

$\frac{1}{x^{2} x - 3 x^{2}}$

Step 7.1.1.2

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

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

Step 7.1.1.3

Factor $x^{2}$ out of $x^{2} x + x^{2} \cdot - 3$ .

$\frac{1}{x^{2} (x - 3)}$

Step 7.1.2

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 $C$ .

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

Step 7.1.3

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

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

Step 7.1.4

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

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Step 7.1.4.1

Cancel the common factor.

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

Step 7.1.4.2

Rewrite the expression.

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

Step 7.1.5

Cancel the common factor of $x - 3$ .

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Step 7.1.5.1

Cancel the common factor.

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

Step 7.1.5.2

Rewrite the expression.

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

Step 7.1.6

Simplify each term.

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Step 7.1.6.1

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

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Step 7.1.6.1.1

Cancel the common factor.

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

Step 7.1.6.1.2

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

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

Step 7.1.6.2

Apply the distributive property.

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

Step 7.1.6.3

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

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

Step 7.1.6.4

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

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Step 7.1.6.4.1

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

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

Step 7.1.6.4.2

Cancel the common factors.

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Step 7.1.6.4.2.1

Raise $x$ to the power of $1$ .

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

Step 7.1.6.4.2.2

Factor $x$ out of $x^{1}$ .

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

Step 7.1.6.4.2.3

Cancel the common factor.

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

Step 7.1.6.4.2.4

Rewrite the expression.

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

Step 7.1.6.4.2.5

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

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

Step 7.1.6.5

Apply the distributive property.

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

Step 7.1.6.6

Multiply $x$ by $x$ .

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

Step 7.1.6.7

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

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

Step 7.1.6.8

Apply the distributive property.

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

Step 7.1.6.9

Rewrite using the commutative property of multiplication.

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

Step 7.1.6.10

Cancel the common factor of $x - 3$ .

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Step 7.1.6.10.1

Cancel the common factor.

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

Step 7.1.6.10.2

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

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

$1 = A x - 3 A + B x^{2} - 3 B x + C x^{2}$

Step 7.1.7

Simplify the expression.

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Step 7.1.7.1

Move $B$ .

$1 = A x - 3 A + B x^{2} - 3 x B + C x^{2}$

Step 7.1.7.2

Move $- 3 A$ .

$1 = A x + B x^{2} - 3 x B + C x^{2} - 3 A$

Step 7.1.7.3

Move $- 3 x B$ .

$1 = A x + B x^{2} + C x^{2} - 3 x B - 3 A$

Step 7.1.7.4

Move $A x$ .

$1 = B x^{2} + C x^{2} + A x - 3 x B - 3 A$

Step 7.2

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

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Step 7.2.1

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.

$0 = B + C$

Step 7.2.2

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.

$0 = A - 3 B$

Step 7.2.3

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.

$1 = - 3 A$

Step 7.2.4

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

$0 = B + C$

$0 = A - 3 B$

$1 = - 3 A$

$0 = B + C$

$0 = A - 3 B$

$1 = - 3 A$

Step 7.3

Solve the system of equations.

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Step 7.3.1

Solve for $A$ in $1 = - 3 A$ .

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Step 7.3.1.1

Rewrite the equation as $- 3 A = 1$ .

$- 3 A = 1$

$0 = B + C$

$0 = A - 3 B$

Step 7.3.1.2

Divide each term in $- 3 A = 1$ by $- 3$ and simplify.

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Step 7.3.1.2.1

Divide each term in $- 3 A = 1$ by $- 3$ .

$\frac{- 3 A}{- 3} = \frac{1}{- 3}$

$0 = B + C$

$0 = A - 3 B$

Step 7.3.1.2.2

Simplify the left side.

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Step 7.3.1.2.2.1

Cancel the common factor of $- 3$ .

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Step 7.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 3 A}{- 3} = \frac{1}{- 3}$

$0 = B + C$

$0 = A - 3 B$

Step 7.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{- 3}$

$0 = B + C$

$0 = A - 3 B$

$A = \frac{1}{- 3}$

$0 = B + C$

$0 = A - 3 B$

$A = \frac{1}{- 3}$

$0 = B + C$

$0 = A - 3 B$

Step 7.3.1.2.3

Simplify the right side.

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Step 7.3.1.2.3.1

Move the negative in front of the fraction.

$A = - \frac{1}{3}$

$0 = B + C$

$0 = A - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

$0 = A - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

$0 = A - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

$0 = A - 3 B$

Step 7.3.2

Replace all occurrences of $A$ with $- \frac{1}{3}$ in each equation.

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Step 7.3.2.1

Replace all occurrences of $A$ in $0 = A - 3 B$ with $- \frac{1}{3}$ .

$0 = (- \frac{1}{3}) - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.2.2

Simplify the right side.

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Step 7.3.2.2.1

Remove parentheses.

$0 = - \frac{1}{3} - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

$0 = - \frac{1}{3} - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

$0 = - \frac{1}{3} - 3 B$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3

Solve for $B$ in $0 = - \frac{1}{3} - 3 B$ .

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Step 7.3.3.1

Rewrite the equation as $- \frac{1}{3} - 3 B = 0$ .

$- \frac{1}{3} - 3 B = 0$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.2

Add $\frac{1}{3}$ to both sides of the equation.

$- 3 B = \frac{1}{3}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3

Divide each term in $- 3 B = \frac{1}{3}$ by $- 3$ and simplify.

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Step 7.3.3.3.1

Divide each term in $- 3 B = \frac{1}{3}$ by $- 3$ .

$\frac{- 3 B}{- 3} = \frac{\frac{1}{3}}{- 3}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3.2

Simplify the left side.

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Step 7.3.3.3.2.1

Cancel the common factor of $- 3$ .

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Step 7.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 3 B}{- 3} = \frac{\frac{1}{3}}{- 3}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{\frac{1}{3}}{- 3}$

$A = - \frac{1}{3}$

$0 = B + C$

$B = \frac{\frac{1}{3}}{- 3}$

$A = - \frac{1}{3}$

$0 = B + C$

$B = \frac{\frac{1}{3}}{- 3}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3.3

Simplify the right side.

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Step 7.3.3.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$B = \frac{1}{3} \cdot \frac{1}{- 3}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3.3.2

Move the negative in front of the fraction.

$B = \frac{1}{3} \cdot (- \frac{1}{3})$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3.3.3

Multiply $\frac{1}{3} (- \frac{1}{3})$ .

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Step 7.3.3.3.3.3.1

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

$B = - \frac{1}{3 \cdot 3}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.3.3.3.3.2

Multiply $3$ by $3$ .

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = B + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = B + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = B + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = B + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = B + C$

Step 7.3.4

Replace all occurrences of $B$ with $- \frac{1}{9}$ in each equation.

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Step 7.3.4.1

Replace all occurrences of $B$ in $0 = B + C$ with $- \frac{1}{9}$ .

$0 = (- \frac{1}{9}) + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

Step 7.3.4.2

Simplify the right side.

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Step 7.3.4.2.1

Remove parentheses.

$0 = - \frac{1}{9} + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = - \frac{1}{9} + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$0 = - \frac{1}{9} + C$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

Step 7.3.5

Solve for $C$ in $0 = - \frac{1}{9} + C$ .

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Step 7.3.5.1

Rewrite the equation as $- \frac{1}{9} + C = 0$ .

$- \frac{1}{9} + C = 0$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

Step 7.3.5.2

Add $\frac{1}{9}$ to both sides of the equation.

$C = \frac{1}{9}$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

$C = \frac{1}{9}$

$B = - \frac{1}{9}$

$A = - \frac{1}{3}$

Step 7.3.6

Solve the system of equations.

$C = \frac{1}{9} B = - \frac{1}{9} A = - \frac{1}{3}$

Step 7.3.7

List all of the solutions.

$C = \frac{1}{9}, B = - \frac{1}{9}, A = - \frac{1}{3}$

Step 7.4

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

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

Step 7.5

Simplify.

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Step 7.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{3} \cdot \frac{1}{x^{2}} + \frac{- \frac{1}{9}}{x} + \frac{\frac{1}{9}}{x - 3}$

Step 7.5.2

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

$- \frac{1}{x^{2} \cdot 3} + \frac{- \frac{1}{9}}{x} + \frac{\frac{1}{9}}{x - 3}$

Step 7.5.3

Move $3$ to the left of $x^{2}$ .

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

Step 7.5.4

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{3 x^{2}} - \frac{1}{9} \cdot \frac{1}{x} + \frac{\frac{1}{9}}{x - 3}$

Step 7.5.5

Multiply $\frac{1}{x}$ by $\frac{1}{9}$ .

$- \frac{1}{3 x^{2}} - \frac{1}{x \cdot 9} + \frac{\frac{1}{9}}{x - 3}$

Step 7.5.6

Move $9$ to the left of $x$ .

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

Step 7.5.7

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{3 x^{2}} - \frac{1}{9 x} + \frac{1}{9} \cdot \frac{1}{x - 3}$

Step 7.5.8

Multiply $\frac{1}{9}$ by $\frac{1}{x - 3}$ .

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

Step 8

Split the single integral into multiple integrals.

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

Step 9

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

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

Step 10

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

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

Step 11

Apply basic rules of exponents.

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Step 11.1

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

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

Step 11.2

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

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Step 11.2.1

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

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

Step 11.2.2

Multiply $2$ by $- 1$ .

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

Step 12

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

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

Step 13

Combine $- x^{- 1}]_{4}^{5}$ and $\frac{1}{3}$ .

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

Step 14

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

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

Step 15

Since $\frac{1}{9}$ is constant with respect to $x$ , move $\frac{1}{9}$ out of the integral.

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

Step 16

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

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

Step 17

Combine $\ln (| x |)]_{4}^{5}$ and $\frac{1}{9}$ .

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

Step 18

Since $\frac{1}{9}$ is constant with respect to $x$ , move $\frac{1}{9}$ out of the integral.

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

Step 19

Let $u = x - 3$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 19.1

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

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Step 19.1.1

Differentiate $x - 3$ .

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

Step 19.1.2

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 19.1.3

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 19.1.4

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

$1 + 0$

Step 19.1.5

Add $1$ and $0$ .

$1$

Step 19.2

Substitute the lower limit in for $x$ in $u = x - 3$ .

$u_{lower} = 4 - 3$

Step 19.3

Subtract $3$ from $4$ .

$u_{lower} = 1$

Step 19.4

Substitute the upper limit in for $x$ in $u = x - 3$ .

$u_{upper} = 5 - 3$

Step 19.5

Subtract $3$ from $5$ .

$u_{upper} = 2$

Step 19.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 19.7

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

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

Step 20

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

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

Step 21

Combine $\frac{1}{9}$ and $\ln (| u |)]_{1}^{2}$ .

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

Step 22

Substitute and simplify.

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Step 22.1

Evaluate $x$ at $5$ and at $4$ .

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

Step 22.2

Evaluate $- x^{- 1}$ at $5$ and at $4$ .

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

Step 22.3

Evaluate $\ln (| x |)$ at $5$ and at $4$ .

$5 - 4 - 9 (- \frac{- 5^{- 1} + 4^{- 1}}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{\ln (| u |)]_{1}^{2}}{9})$

Step 22.4

Evaluate $\ln (| u |)$ at $2$ and at $1$ .

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

Step 22.5

Simplify.

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Step 22.5.1

Subtract $4$ from $5$ .

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

Step 22.5.2

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

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

Step 22.5.3

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

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

Step 22.5.4

To write $- \frac{1}{5}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 22.5.5

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

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

Step 22.5.6

Write each expression with a common denominator of $20$ , by multiplying each by an appropriate factor of $1$ .

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Step 22.5.6.1

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

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

Step 22.5.6.2

Multiply $5$ by $4$ .

$1 - 9 (- \frac{- \frac{4}{20} + \frac{1}{4} \cdot \frac{5}{5}}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.6.3

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

$1 - 9 (- \frac{- \frac{4}{20} + \frac{5}{4 \cdot 5}}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.6.4

Multiply $4$ by $5$ .

$1 - 9 (- \frac{- \frac{4}{20} + \frac{5}{20}}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.7

Combine the numerators over the common denominator.

$1 - 9 (- \frac{\frac{- 4 + 5}{20}}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.8

Add $- 4$ and $5$ .

$1 - 9 (- \frac{\frac{1}{20}}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.9

Rewrite $\frac{\frac{1}{20}}{3}$ as a product.

$1 - 9 (- (\frac{1}{20} \cdot \frac{1}{3}) - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.10

Multiply $\frac{1}{20}$ by $\frac{1}{3}$ .

$1 - 9 (- \frac{1}{20 \cdot 3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.11

Multiply $20$ by $3$ .

$1 - 9 (- \frac{1}{60} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.12

To write $- \frac{1}{60}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$1 - 9 (- \frac{1}{60} \cdot \frac{3}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.13

To write $- \frac{\ln (| 5 |) - \ln (| 4 |)}{9}$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$1 - 9 (- \frac{1}{60} \cdot \frac{3}{3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} \cdot \frac{20}{20} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.14

Write each expression with a common denominator of $180$ , by multiplying each by an appropriate factor of $1$ .

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Step 22.5.14.1

Multiply $\frac{1}{60}$ by $\frac{3}{3}$ .

$1 - 9 (- \frac{3}{60 \cdot 3} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} \cdot \frac{20}{20} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.14.2

Multiply $60$ by $3$ .

$1 - 9 (- \frac{3}{180} - \frac{\ln (| 5 |) - \ln (| 4 |)}{9} \cdot \frac{20}{20} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.14.3

Multiply $\frac{\ln (| 5 |) - \ln (| 4 |)}{9}$ by $\frac{20}{20}$ .

$1 - 9 (- \frac{3}{180} - \frac{(\ln (| 5 |) - \ln (| 4 |)) \cdot 20}{9 \cdot 20} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.14.4

Multiply $9$ by $20$ .

$1 - 9 (- \frac{3}{180} - \frac{(\ln (| 5 |) - \ln (| 4 |)) \cdot 20}{180} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.15

Combine the numerators over the common denominator.

$1 - 9 (\frac{- 3 - (\ln (| 5 |) - \ln (| 4 |)) \cdot 20}{180} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.16

Multiply $20$ by $- 1$ .

$1 - 9 (\frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |))}{180} + \frac{(\ln (| 2 |)) - \ln (| 1 |)}{9})$

Step 22.5.17

To write $\frac{\ln (| 2 |) - \ln (| 1 |)}{9}$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$1 - 9 (\frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |))}{180} + \frac{\ln (| 2 |) - \ln (| 1 |)}{9} \cdot \frac{20}{20})$

Step 22.5.18

Write each expression with a common denominator of $180$ , by multiplying each by an appropriate factor of $1$ .

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Step 22.5.18.1

Multiply $\frac{\ln (| 2 |) - \ln (| 1 |)}{9}$ by $\frac{20}{20}$ .

$1 - 9 (\frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |))}{180} + \frac{(\ln (| 2 |) - \ln (| 1 |)) \cdot 20}{9 \cdot 20})$

Step 22.5.18.2

Multiply $9$ by $20$ .

$1 - 9 (\frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |))}{180} + \frac{(\ln (| 2 |) - \ln (| 1 |)) \cdot 20}{180})$

Step 22.5.19

Combine the numerators over the common denominator.

$1 - 9 \frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + (\ln (| 2 |) - \ln (| 1 |)) \cdot 20}{180}$

Step 22.5.20

Move $20$ to the left of $\ln (| 2 |) - \ln (| 1 |)$ .

$1 - 9 \frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 \cdot (\ln (| 2 |) - \ln (| 1 |))}{180}$

Step 22.5.21

Combine $- 9$ and $\frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |))}{180}$ .

$1 + \frac{- 9 (- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |)))}{180}$

Step 22.5.22

Cancel the common factor of $- 9$ and $180$ .

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Step 22.5.22.1

Factor $9$ out of $- 9 (- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |)))$ .

$1 + \frac{9 (- (- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |))))}{180}$

Step 22.5.22.2

Cancel the common factors.

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Step 22.5.22.2.1

Factor $9$ out of $180$ .

$1 + \frac{9 (- (- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |))))}{9 (20)}$

Step 22.5.22.2.2

Cancel the common factor.

$1 + \frac{9 (- (- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |))))}{9 \cdot 20}$

Step 22.5.22.2.3

Rewrite the expression.

$1 + \frac{- (- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |)))}{20}$

Step 22.5.23

Move the negative in front of the fraction.

$1 - \frac{- 3 - 20 (\ln (| 5 |) - \ln (| 4 |)) + 20 (\ln (| 2 |) - \ln (| 1 |))}{20}$

Step 23

Simplify.

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Step 23.1

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

$1 - \frac{- 3 - 20 \ln (\frac{| 5 |}{| 4 |}) + 20 (\ln (| 2 |) - \ln (| 1 |))}{20}$

Step 23.2

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

$1 - \frac{- 3 - 20 \ln (\frac{| 5 |}{| 4 |}) + 20 \ln (\frac{| 2 |}{| 1 |})}{20}$

Step 23.3

Write $1$ as a fraction with a common denominator.

$\frac{20}{20} - \frac{- 3 - 20 \ln (\frac{| 5 |}{| 4 |}) + 20 \ln (\frac{| 2 |}{| 1 |})}{20}$

Step 23.4

Combine the numerators over the common denominator.

$\frac{20 - (- 3 - 20 \ln (\frac{| 5 |}{| 4 |}) + 20 \ln (\frac{| 2 |}{| 1 |}))}{20}$

Step 24

Simplify.

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Step 24.1

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$\frac{20 - (- 3 - 20 \ln (\frac{5}{| 4 |}) + 20 \ln (\frac{| 2 |}{| 1 |}))}{20}$

Step 24.2

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$\frac{20 - (- 3 - 20 \ln (\frac{5}{4}) + 20 \ln (\frac{| 2 |}{| 1 |}))}{20}$

Step 24.3

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

$\frac{20 - (- 3 - 20 \ln (\frac{5}{4}) + 20 \ln (\frac{2}{| 1 |}))}{20}$

Step 24.4

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

$\frac{20 - (- 3 - 20 \ln (\frac{5}{4}) + 20 \ln (\frac{2}{1}))}{20}$

Step 24.5

Divide $2$ by $1$ .

$\frac{20 - (- 3 - 20 \ln (\frac{5}{4}) + 20 \ln (2))}{20}$

Step 24.6

Apply the distributive property.

$\frac{20 - - 3 - (- 20 \ln (\frac{5}{4})) - (20 \ln (2))}{20}$

Step 24.7

Simplify.

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Step 24.7.1

Multiply $- 1$ by $- 3$ .

$\frac{20 + 3 - (- 20 \ln (\frac{5}{4})) - (20 \ln (2))}{20}$

Step 24.7.2

Multiply $- 20$ by $- 1$ .

$\frac{20 + 3 + 20 \ln (\frac{5}{4}) - (20 \ln (2))}{20}$

Step 24.7.3

Multiply $20$ by $- 1$ .

$\frac{20 + 3 + 20 \ln (\frac{5}{4}) - 20 \ln (2)}{20}$

Step 24.8

Add $20$ and $3$ .

$\frac{23 + 20 \ln (\frac{5}{4}) - 20 \ln (2)}{20}$

Step 25

The result can be shown in multiple forms.

Exact Form:

$\frac{23 + 20 \ln (\frac{5}{4}) - 20 \ln (2)}{20}$

Decimal Form:

$0.67999637 \dots$

Step 26