Calculus Examples

$\frac{- x^{3} + 17 x^{2} - 12 x - 9}{x^{4} - 3 x^{3}}$

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 $x^{3}$ out of $x^{4} - 3 x^{3}$ .

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Step 1.1.1.1

Factor $x^{3}$ out of $x^{4}$ .

$\frac{- x^{3} + 17 x^{2} - 12 x - 9}{x^{3} x - 3 x^{3}}$

Step 1.1.1.2

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

$\frac{- x^{3} + 17 x^{2} - 12 x - 9}{x^{3} x + x^{3} \cdot - 3}$

Step 1.1.1.3

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

$\frac{- x^{3} + 17 x^{2} - 12 x - 9}{x^{3} (x - 3)}$

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 $D$ .

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

Step 1.1.3

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

$\frac{(- x^{3} + 17 x^{2} - 12 x - 9) (x^{3} (x - 3))}{x^{3} (x - 3)} = \frac{A (x^{3} (x - 3))}{x^{3}} + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.4

Cancel the common factor of $x^{3}$ .

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Step 1.1.4.1

Cancel the common factor.

Step 1.1.4.2

Rewrite the expression.

$\frac{(- x^{3} + 17 x^{2} - 12 x - 9) (x - 3)}{x - 3} = \frac{A (x^{3} (x - 3))}{x^{3}} + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.5

Cancel the common factor of $x - 3$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(- x^{3} + 17 x^{2} - 12 x - 9) (x - 3)}{x - 3} = \frac{A (x^{3} (x - 3))}{x^{3}} + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.5.2

Divide $- x^{3} + 17 x^{2} - 12 x - 9$ by $1$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = \frac{A (x^{3} (x - 3))}{x^{3}} + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $x^{3}$ .

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Step 1.1.6.1.1

Cancel the common factor.

$- x^{3} + 17 x^{2} - 12 x - 9 = \frac{A (x^{3} (x - 3))}{x^{3}} + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.1.2

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A (x - 3) + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.2

Apply the distributive property.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x + A \cdot - 3 + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.3

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 \cdot A + \frac{B (x^{3} (x - 3))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.4

Cancel the common factor of $x^{3}$ and $x^{2}$ .

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Step 1.1.6.4.1

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + \frac{x^{2} (B (x (x - 3)))}{x^{2}} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.4.2

Cancel the common factors.

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Step 1.1.6.4.2.1

Multiply by $1$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + \frac{x^{2} (B (x (x - 3)))}{x^{2} \cdot 1} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.4.2.2

Cancel the common factor.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + \frac{x^{2} (B (x (x - 3)))}{x^{2} \cdot 1} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.4.2.3

Rewrite the expression.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + \frac{B (x (x - 3))}{1} + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.4.2.4

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B (x (x - 3)) + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.5

Apply the distributive property.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B (x \cdot x + x \cdot - 3) + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.6

Multiply $x$ by $x$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B (x^{2} + x \cdot - 3) + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.7

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B (x^{2} - 3 \cdot x) + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.8

Apply the distributive property.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} + B (- 3 x) + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.9

Rewrite using the commutative property of multiplication.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + \frac{C (x^{3} (x - 3))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.10

Cancel the common factor of $x^{3}$ and $x$ .

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Step 1.1.6.10.1

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + \frac{x (C (x^{2} (x - 3)))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.10.2

Cancel the common factors.

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Step 1.1.6.10.2.1

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + \frac{x (C (x^{2} (x - 3)))}{x} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.10.2.2

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + \frac{x (C (x^{2} (x - 3)))}{x \cdot 1} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.10.2.3

Cancel the common factor.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + \frac{x (C (x^{2} (x - 3)))}{x \cdot 1} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.10.2.4

Rewrite the expression.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + \frac{C (x^{2} (x - 3))}{1} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.10.2.5

Divide $C (x^{2} (x - 3))$ by $1$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C (x^{2} (x - 3)) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.11

Apply the distributive property.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C (x^{2} x + x^{2} \cdot - 3) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.12

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.6.12.1

Multiply $x^{2}$ by $x$ .

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Step 1.1.6.12.1.1

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C (x^{2} x + x^{2} \cdot - 3) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.12.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C (x^{2 + 1} + x^{2} \cdot - 3) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.12.2

Add $2$ and $1$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C (x^{3} + x^{2} \cdot - 3) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.13

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

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C (x^{3} - 3 \cdot x^{2}) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.14

Apply the distributive property.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C x^{3} + C (- 3 x^{2}) + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.15

Rewrite using the commutative property of multiplication.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C x^{3} - 3 C x^{2} + \frac{(D) (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.16

Cancel the common factor of $x - 3$ .

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Step 1.1.6.16.1

Cancel the common factor.

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C x^{3} - 3 C x^{2} + \frac{D (x^{3} (x - 3))}{x - 3}$

Step 1.1.6.16.2

Divide $(D) (x^{3})$ by $1$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C x^{3} - 3 C x^{2} + (D) (x^{3})$

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 B x + C x^{3} - 3 C x^{2} + D x^{3}$

Step 1.1.7

Simplify the expression.

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Step 1.1.7.1

Move $B$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x - 3 A + B x^{2} - 3 x B + C x^{3} - 3 C x^{2} + D x^{3}$

Step 1.1.7.2

Move $- 3 A$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x + B x^{2} - 3 x B + C x^{3} - 3 C x^{2} + D x^{3} - 3 A$

Step 1.1.7.3

Move $- 3 x B$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = A x + B x^{2} + C x^{3} - 3 C x^{2} + D x^{3} - 3 x B - 3 A$

Step 1.1.7.4

Move $A x$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = B x^{2} + C x^{3} - 3 C x^{2} + D x^{3} + A x - 3 x B - 3 A$

Step 1.1.7.5

Move $- 3 C x^{2}$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = B x^{2} + C x^{3} + D x^{3} - 3 C x^{2} + A x - 3 x B - 3 A$

Step 1.1.7.6

Move $B x^{2}$ .

$- x^{3} + 17 x^{2} - 12 x - 9 = C x^{3} + D x^{3} + B x^{2} - 3 C x^{2} + A x - 3 x B - 3 A$

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 $x^{3}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$- 1 = C + D$

Step 1.2.2

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.

$17 = B - 3 C$

Step 1.2.3

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.

$- 12 = A - 3 B$

Step 1.2.4

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.

$- 9 = - 3 A$

Step 1.2.5

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

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$- 9 = - 3 A$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$- 9 = - 3 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $- 9 = - 3 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 3 A = - 9$ .

$- 3 A = - 9$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

Step 1.3.1.2

Divide each term in $- 3 A = - 9$ by $- 3$ and simplify.

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Step 1.3.1.2.1

Divide each term in $- 3 A = - 9$ by $- 3$ .

$\frac{- 3 A}{- 3} = \frac{- 9}{- 3}$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Cancel the common factor of $- 3$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 3 A}{- 3} = \frac{- 9}{- 3}$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 9}{- 3}$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$A = \frac{- 9}{- 3}$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$A = \frac{- 9}{- 3}$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $- 9$ by $- 3$ .

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = A - 3 B$

Step 1.3.2

Replace all occurrences of $A$ with $3$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $- 12 = A - 3 B$ with $3$ .

$- 12 = (3) - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$- 12 = 3 - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = 3 - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 12 = 3 - 3 B$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3

Solve for $B$ in $- 12 = 3 - 3 B$ .

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Step 1.3.3.1

Rewrite the equation as $3 - 3 B = - 12$ .

$3 - 3 B = - 12$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Subtract $3$ from both sides of the equation.

$- 3 B = - 12 - 3$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3.2.2

Subtract $3$ from $- 12$ .

$- 3 B = - 15$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$- 3 B = - 15$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3.3

Divide each term in $- 3 B = - 15$ by $- 3$ and simplify.

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Step 1.3.3.3.1

Divide each term in $- 3 B = - 15$ by $- 3$ .

$\frac{- 3 B}{- 3} = \frac{- 15}{- 3}$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $- 3$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 3 B}{- 3} = \frac{- 15}{- 3}$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 15}{- 3}$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$B = \frac{- 15}{- 3}$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$B = \frac{- 15}{- 3}$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $- 15$ by $- 3$ .

$B = 5$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

$17 = B - 3 C$

Step 1.3.4

Replace all occurrences of $B$ with $5$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $17 = B - 3 C$ with $5$ .

$17 = (5) - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Remove parentheses.

$17 = 5 - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

$17 = 5 - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

$17 = 5 - 3 C$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5

Solve for $C$ in $17 = 5 - 3 C$ .

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Step 1.3.5.1

Rewrite the equation as $5 - 3 C = 17$ .

$5 - 3 C = 17$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5.2

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

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Step 1.3.5.2.1

Subtract $5$ from both sides of the equation.

$- 3 C = 17 - 5$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5.2.2

Subtract $5$ from $17$ .

$- 3 C = 12$

$B = 5$

$A = 3$

$- 1 = C + D$

$- 3 C = 12$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5.3

Divide each term in $- 3 C = 12$ by $- 3$ and simplify.

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Step 1.3.5.3.1

Divide each term in $- 3 C = 12$ by $- 3$ .

$\frac{- 3 C}{- 3} = \frac{12}{- 3}$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5.3.2

Simplify the left side.

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Step 1.3.5.3.2.1

Cancel the common factor of $- 3$ .

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Step 1.3.5.3.2.1.1

Cancel the common factor.

$\frac{- 3 C}{- 3} = \frac{12}{- 3}$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{12}{- 3}$

$B = 5$

$A = 3$

$- 1 = C + D$

$C = \frac{12}{- 3}$

$B = 5$

$A = 3$

$- 1 = C + D$

$C = \frac{12}{- 3}$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.5.3.3

Simplify the right side.

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Step 1.3.5.3.3.1

Divide $12$ by $- 3$ .

$C = - 4$

$B = 5$

$A = 3$

$- 1 = C + D$

$C = - 4$

$B = 5$

$A = 3$

$- 1 = C + D$

$C = - 4$

$B = 5$

$A = 3$

$- 1 = C + D$

$C = - 4$

$B = 5$

$A = 3$

$- 1 = C + D$

Step 1.3.6

Replace all occurrences of $C$ with $- 4$ in each equation.

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Step 1.3.6.1

Replace all occurrences of $C$ in $- 1 = C + D$ with $- 4$ .

$- 1 = (- 4) + D$

$C = - 4$

$B = 5$

$A = 3$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Remove parentheses.

$- 1 = - 4 + D$

$C = - 4$

$B = 5$

$A = 3$

$- 1 = - 4 + D$

$C = - 4$

$B = 5$

$A = 3$

$- 1 = - 4 + D$

$C = - 4$

$B = 5$

$A = 3$

Step 1.3.7

Solve for $D$ in $- 1 = - 4 + D$ .

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Step 1.3.7.1

Rewrite the equation as $- 4 + D = - 1$ .

$- 4 + D = - 1$

$C = - 4$

$B = 5$

$A = 3$

Step 1.3.7.2

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

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Step 1.3.7.2.1

Add $4$ to both sides of the equation.

$D = - 1 + 4$

$C = - 4$

$B = 5$

$A = 3$

Step 1.3.7.2.2

Add $- 1$ and $4$ .

$D = 3$

$C = - 4$

$B = 5$

$A = 3$

$D = 3$

$C = - 4$

$B = 5$

$A = 3$

$D = 3$

$C = - 4$

$B = 5$

$A = 3$

Step 1.3.8

Solve the system of equations.

$D = 3 C = - 4 B = 5 A = 3$

Step 1.3.9

List all of the solutions.

$D = 3, C = - 4, B = 5, A = 3$

Step 1.4

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

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

Step 1.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

$3 \int \frac{1}{x^{3}} d x + \int \frac{5}{x^{2}} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 4

Apply basic rules of exponents.

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Step 4.1

Move $x^{3}$ out of the denominator by raising it to the $- 1$ power.

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

Step 4.2

Multiply the exponents in ${(x^{3})}^{- 1}$ .

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Step 4.2.1

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

$3 \int x^{3 \cdot - 1} d x + \int \frac{5}{x^{2}} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 4.2.2

Multiply $3$ by $- 1$ .

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

Step 5

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

$3 (- \frac{1}{2} x^{- 2} + C) + \int \frac{5}{x^{2}} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 6

Simplify.

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Step 6.1

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

$3 (- \frac{x^{- 2}}{2} + C) + \int \frac{5}{x^{2}} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 6.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$3 (- \frac{1}{2 x^{2}} + C) + \int \frac{5}{x^{2}} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 7

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$3 (- \frac{1}{2 x^{2}} + C) + 5 \int \frac{1}{x^{2}} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 8

Apply basic rules of exponents.

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Step 8.1

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 \int {(x^{2})}^{- 1} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 8.2

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

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Step 8.2.1

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 \int x^{2 \cdot - 1} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 8.2.2

Multiply $2$ by $- 1$ .

$3 (- \frac{1}{2 x^{2}} + C) + 5 \int x^{- 2} d x + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 9

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) + \int - \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 10

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) - \int \frac{4}{x} d x + \int \frac{3}{x - 3} d x$

Step 11

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) - (4 \int \frac{1}{x} d x) + \int \frac{3}{x - 3} d x$

Step 12

Multiply $4$ by $- 1$ .

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) - 4 \int \frac{1}{x} d x + \int \frac{3}{x - 3} d x$

Step 13

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) - 4 (\ln (| x |) + C) + \int \frac{3}{x - 3} d x$

Step 14

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

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) - 4 (\ln (| x |) + C) + 3 \int \frac{1}{x - 3} d x$

Step 15

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

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Step 15.1

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

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Step 15.1.1

Differentiate $x - 3$ .

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

Step 15.1.2

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 15.1.3

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 15.1.4

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

$1 + 0$

Step 15.1.5

Add $1$ and $0$ .

$1$

Step 15.2

Rewrite the problem using $u$ and $d u$ .

$3 (- \frac{1}{2 x^{2}} + C) + 5 (- x^{- 1} + C) - 4 (\ln (| x |) + C) + 3 \int \frac{1}{u} d u$

Step 16

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

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

Step 17

Simplify.

$- \frac{3}{2 x^{2}} - \frac{5}{x} - 4 \ln (| x |) + 3 \ln (| u |) + C$

Step 18

Replace all occurrences of $u$ with $x - 3$ .

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