Precalculus Examples

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

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

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

Step 1.2

$\frac{A}{x} + \frac{B x + C}{{(2 x^{2} + 1)}^{2}} + \frac{D x + F}{2 x^{2} + 1}$

Step 1.3

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

$\frac{(4 x - 5) (x {(2 x^{2} + 1)}^{2})}{x {(2 x^{2} + 1)}^{2}} = \frac{A (x {(2 x^{2} + 1)}^{2})}{x} + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

Cancel the common factor.

Step 1.4.2

Rewrite the expression.

$\frac{(4 x - 5) ({(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} = \frac{A (x {(2 x^{2} + 1)}^{2})}{x} + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.5

Cancel the common factor of ${(2 x^{2} + 1)}^{2}$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(4 x - 5) {(2 x^{2} + 1)}^{2}}{{(2 x^{2} + 1)}^{2}} = \frac{A (x {(2 x^{2} + 1)}^{2})}{x} + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.5.2

Divide $4 x - 5$ by $1$ .

$4 x - 5 = \frac{A (x {(2 x^{2} + 1)}^{2})}{x} + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x$ .

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Step 1.6.1.1

Cancel the common factor.

$4 x - 5 = \frac{A (x {(2 x^{2} + 1)}^{2})}{x} + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.1.2

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

$4 x - 5 = A ({(2 x^{2} + 1)}^{2}) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.2

Rewrite ${(2 x^{2} + 1)}^{2}$ as $(2 x^{2} + 1) (2 x^{2} + 1)$ .

$4 x - 5 = A ((2 x^{2} + 1) (2 x^{2} + 1)) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.3

Expand $(2 x^{2} + 1) (2 x^{2} + 1)$ using the FOIL Method.

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Step 1.6.3.1

Apply the distributive property.

$4 x - 5 = A (2 x^{2} (2 x^{2} + 1) + 1 (2 x^{2} + 1)) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.3.2

Apply the distributive property.

$4 x - 5 = A (2 x^{2} (2 x^{2}) + 2 x^{2} \cdot 1 + 1 (2 x^{2} + 1)) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.3.3

Apply the distributive property.

$4 x - 5 = A (2 x^{2} (2 x^{2}) + 2 x^{2} \cdot 1 + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4

Simplify and combine like terms.

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Step 1.6.4.1

Simplify each term.

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Step 1.6.4.1.1

Rewrite using the commutative property of multiplication.

$4 x - 5 = A (2 \cdot (2 x^{2} x^{2}) + 2 x^{2} \cdot 1 + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.2

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

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Step 1.6.4.1.2.1

Move $x^{2}$ .

$4 x - 5 = A (2 \cdot (2 (x^{2} x^{2})) + 2 x^{2} \cdot 1 + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.2.2

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

$4 x - 5 = A (2 \cdot (2 x^{2 + 2}) + 2 x^{2} \cdot 1 + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.2.3

Add $2$ and $2$ .

$4 x - 5 = A (2 \cdot (2 x^{4}) + 2 x^{2} \cdot 1 + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.3

Multiply $2$ by $2$ .

$4 x - 5 = A (4 x^{4} + 2 x^{2} \cdot 1 + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.4

Multiply $2$ by $1$ .

$4 x - 5 = A (4 x^{4} + 2 x^{2} + 1 (2 x^{2}) + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.5

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

$4 x - 5 = A (4 x^{4} + 2 x^{2} + 2 x^{2} + 1 \cdot 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.1.6

Multiply $1$ by $1$ .

$4 x - 5 = A (4 x^{4} + 2 x^{2} + 2 x^{2} + 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.4.2

Add $2 x^{2}$ and $2 x^{2}$ .

$4 x - 5 = A (4 x^{4} + 4 x^{2} + 1) + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.5

Apply the distributive property.

$4 x - 5 = A (4 x^{4}) + A (4 x^{2}) + A \cdot 1 + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.6

Simplify.

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Step 1.6.6.1

Rewrite using the commutative property of multiplication.

$4 x - 5 = 4 A x^{4} + A (4 x^{2}) + A \cdot 1 + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.6.2

Rewrite using the commutative property of multiplication.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A \cdot 1 + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.6.3

Multiply $A$ by $1$ .

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.7

Cancel the common factor of ${(2 x^{2} + 1)}^{2}$ .

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Step 1.6.7.1

Cancel the common factor.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + \frac{(B x + C) (x {(2 x^{2} + 1)}^{2})}{{(2 x^{2} + 1)}^{2}} + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.7.2

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

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + (B x + C) (x) + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.8

Apply the distributive property.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x \cdot x + C x + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.9

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

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Step 1.6.9.1

Move $x$ .

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B (x \cdot x) + C x + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.9.2

Multiply $x$ by $x$ .

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + \frac{(D x + F) (x {(2 x^{2} + 1)}^{2})}{2 x^{2} + 1}$

Step 1.6.10

Cancel the common factor of ${(2 x^{2} + 1)}^{2}$ and $2 x^{2} + 1$ .

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Step 1.6.10.1

Factor $2 x^{2} + 1$ out of $(D x + F) (x {(2 x^{2} + 1)}^{2})$ .

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + \frac{(2 x^{2} + 1) ((D x + F) (x (2 x^{2} + 1)))}{2 x^{2} + 1}$

Step 1.6.10.2

Cancel the common factors.

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Step 1.6.10.2.1

Multiply by $1$ .

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + \frac{(2 x^{2} + 1) ((D x + F) (x (2 x^{2} + 1)))}{(2 x^{2} + 1) \cdot 1}$

Step 1.6.10.2.2

Cancel the common factor.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + \frac{(2 x^{2} + 1) ((D x + F) (x (2 x^{2} + 1)))}{(2 x^{2} + 1) \cdot 1}$

Step 1.6.10.2.3

Rewrite the expression.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + \frac{(D x + F) (x (2 x^{2} + 1))}{1}$

Step 1.6.10.2.4

Divide $(D x + F) (x (2 x^{2} + 1))$ by $1$ .

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + (D x + F) (x (2 x^{2} + 1))$

Step 1.6.11

Apply the distributive property.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + (D x + F) (x (2 x^{2}) + x \cdot 1)$

Step 1.6.12

Rewrite using the commutative property of multiplication.

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + (D x + F) (2 x \cdot x^{2} + x \cdot 1)$

Step 1.6.13

Multiply $x$ by $1$ .

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

Step 1.6.14

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

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Step 1.6.14.1

Move $x^{2}$ .

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

Step 1.6.14.2

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

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Step 1.6.14.2.1

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

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

Step 1.6.14.2.2

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

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + (D x + F) (2 x^{2 + 1} + x)$

Step 1.6.14.3

Add $2$ and $1$ .

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

Step 1.6.15

Expand $(D x + F) (2 x^{3} + x)$ using the FOIL Method.

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Step 1.6.15.1

Apply the distributive property.

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

Step 1.6.15.2

Apply the distributive property.

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

Step 1.6.15.3

Apply the distributive property.

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

Step 1.6.16

Simplify each term.

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Step 1.6.16.1

Rewrite using the commutative property of multiplication.

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

Step 1.6.16.2

Multiply $x$ by $x^{3}$ by adding the exponents.

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Step 1.6.16.2.1

Move $x^{3}$ .

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

Step 1.6.16.2.2

Multiply $x^{3}$ by $x$ .

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Step 1.6.16.2.2.1

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

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

Step 1.6.16.2.2.2

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

$4 x - 5 = 4 A x^{4} + 4 A x^{2} + A + B x^{2} + C x + 2 D x^{3 + 1} + D x \cdot x + F (2 x^{3}) + F x$

Step 1.6.16.2.3

Add $3$ and $1$ .

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

Step 1.6.16.3

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

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Step 1.6.16.3.1

Move $x$ .

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

Step 1.6.16.3.2

Multiply $x$ by $x$ .

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

Step 1.6.16.4

Rewrite using the commutative property of multiplication.

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

Step 1.7

Simplify the expression.

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Step 1.7.1

Move $A$ .

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

Step 1.7.2

Move $A$ .

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

Step 1.7.3

Reorder $B$ and $x^{2}$ .

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

Step 1.7.4

Move $F$ .

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

Step 1.7.5

Reorder $F$ and $x$ .

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

Step 1.7.6

Move $A$ .

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

Step 1.7.7

Move $C x$ .

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

Step 1.7.8

Move $D x^{2}$ .

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

Step 1.7.9

Move $B x^{2}$ .

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

Step 1.7.10

Move $4 x^{2} A$ .

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

Step 2

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

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Step 2.1

Create an equation for the partial fraction variables by equating the coefficients of $x^{4}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 4 A + 2 D$

Step 2.2

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.

$0 = 2 F$

Step 2.3

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.

$0 = 4 A + B + D$

Step 2.4

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.

$4 = C + F$

Step 2.5

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.

$- 5 = A$

Step 2.6

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

$0 = 4 A + 2 D$

$0 = 2 F$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

$0 = 4 A + 2 D$

$0 = 2 F$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $F$ in $0 = 2 F$ .

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Step 3.1.1

Rewrite the equation as $2 F = 0$ .

$2 F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

Step 3.1.2

Divide each term in $2 F = 0$ by $2$ and simplify.

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Step 3.1.2.1

Divide each term in $2 F = 0$ by $2$ .

$\frac{2 F}{2} = \frac{0}{2}$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $2$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{2 F}{2} = \frac{0}{2}$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

Step 3.1.2.2.1.2

Divide $F$ by $1$ .

$F = \frac{0}{2}$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

$F = \frac{0}{2}$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

$F = \frac{0}{2}$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $0$ by $2$ .

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$4 = C + F$

$- 5 = A$

Step 3.2

Replace all occurrences of $F$ with $0$ in each equation.

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Step 3.2.1

Replace all occurrences of $F$ in $4 = C + F$ with $0$ .

$4 = C + 0$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$- 5 = A$

Step 3.2.2

Simplify $4 = C + 0$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$4 = C + 0$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$- 5 = A$

$4 = C + 0$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$- 5 = A$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Add $C$ and $0$ .

$4 = C$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$- 5 = A$

$4 = C$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$- 5 = A$

$4 = C$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$- 5 = A$

Step 3.2.3

Rewrite the equation as $A = - 5$ .

$A = - 5$

$4 = C$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

$A = - 5$

$4 = C$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

Step 3.3

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

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Step 3.3.1

Rewrite the equation as $C = 4$ .

$C = 4$

$A = - 5$

$F = 0$

$0 = 4 A + 2 D$

$0 = 4 A + B + D$

Step 3.3.2

Replace all occurrences of $A$ in $0 = 4 A + 2 D$ with $- 5$ .

$0 = 4 (- 5) + 2 D$

$C = 4$

$A = - 5$

$F = 0$

$0 = 4 A + B + D$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Multiply $4$ by $- 5$ .

$0 = - 20 + 2 D$

$C = 4$

$A = - 5$

$F = 0$

$0 = 4 A + B + D$

$0 = - 20 + 2 D$

$C = 4$

$A = - 5$

$F = 0$

$0 = 4 A + B + D$

Step 3.3.4

Replace all occurrences of $A$ in $0 = 4 A + B + D$ with $- 5$ .

$0 = 4 (- 5) + B + D$

$0 = - 20 + 2 D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.3.5

Simplify the right side.

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Step 3.3.5.1

Multiply $4$ by $- 5$ .

$0 = - 20 + B + D$

$0 = - 20 + 2 D$

$C = 4$

$A = - 5$

$F = 0$

$0 = - 20 + B + D$

$0 = - 20 + 2 D$

$C = 4$

$A = - 5$

$F = 0$

$0 = - 20 + B + D$

$0 = - 20 + 2 D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.4

Solve for $D$ in $0 = - 20 + 2 D$ .

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Step 3.4.1

Rewrite the equation as $- 20 + 2 D = 0$ .

$- 20 + 2 D = 0$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.4.2

Add $20$ to both sides of the equation.

$2 D = 20$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.4.3

Divide each term in $2 D = 20$ by $2$ and simplify.

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Step 3.4.3.1

Divide each term in $2 D = 20$ by $2$ .

$\frac{2 D}{2} = \frac{20}{2}$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $2$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

$\frac{2 D}{2} = \frac{20}{2}$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.4.3.2.1.2

Divide $D$ by $1$ .

$D = \frac{20}{2}$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

$D = \frac{20}{2}$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

$D = \frac{20}{2}$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.4.3.3

Simplify the right side.

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Step 3.4.3.3.1

Divide $20$ by $2$ .

$D = 10$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

$D = 10$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

$D = 10$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

$D = 10$

$0 = - 20 + B + D$

$C = 4$

$A = - 5$

$F = 0$

Step 3.5

Replace all occurrences of $D$ with $10$ in each equation.

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Step 3.5.1

Replace all occurrences of $D$ in $0 = - 20 + B + D$ with $10$ .

$0 = - 20 + B + 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

Step 3.5.2

Simplify $0 = - 20 + B + 10$ .

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Step 3.5.2.1

Simplify the left side.

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Step 3.5.2.1.1

Remove parentheses.

$0 = - 20 + B + 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

$0 = - 20 + B + 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

Step 3.5.2.2

Simplify the right side.

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Step 3.5.2.2.1

Add $- 20$ and $10$ .

$0 = B - 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

$0 = B - 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

$0 = B - 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

$0 = B - 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

Step 3.6

Solve for $B$ in $0 = B - 10$ .

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Step 3.6.1

Rewrite the equation as $B - 10 = 0$ .

$B - 10 = 0$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

Step 3.6.2

Add $10$ to both sides of the equation.

$B = 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

$B = 10$

$D = 10$

$C = 4$

$A = - 5$

$F = 0$

Step 3.7

Solve the system of equations.

$B = 10 D = 10 C = 4 A = - 5 F = 0$

Step 3.8

List all of the solutions.

$B = 10, D = 10, C = 4, A = - 5, F = 0$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B x + C}{{(2 x^{2} + 1)}^{2}} + \frac{D x + F}{2 x^{2} + 1}$ with the values found for $A$ , $B$ , $C$ , $D$ , and $F$ .

$- \frac{5}{x} + \frac{10 x + 4}{{(2 x^{2} + 1)}^{2}} + \frac{10 x + 0}{2 x^{2} + 1}$

Step 5

Simplify.

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Step 5.1

Remove parentheses.

$\frac{- 5}{x} + \frac{(10) \cdot x + 4}{{(2 x^{2} + 1)}^{2}} + \frac{(10) \cdot x + 0}{2 x^{2} + 1}$

Step 5.2

Multiply $10$ by $x$ .

$\frac{- 5}{x} + \frac{10 x + 4}{{(2 x^{2} + 1)}^{2}} + \frac{(10) \cdot x + 0}{2 x^{2} + 1}$

Step 5.3

Add $(10) \cdot x$ and $0$ .

$\frac{- 5}{x} + \frac{10 x + 4}{{(2 x^{2} + 1)}^{2}} + \frac{10 x}{2 x^{2} + 1}$