Precalculus Examples

$\frac{x^{2} + 6 x + 1}{{(x^{2} + 4)}^{2}}$

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.

$\frac{A x + B}{{(x^{2} + 4)}^{2}}$

Step 1.2

$\frac{A x + B}{{(x^{2} + 4)}^{2}} + \frac{C x + D}{x^{2} + 4}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x^{2} + 4)}^{2}$ .

$\frac{(x^{2} + 6 x + 1) {(x^{2} + 4)}^{2}}{{(x^{2} + 4)}^{2}} = \frac{(A x + B) {(x^{2} + 4)}^{2}}{{(x^{2} + 4)}^{2}} + \frac{(C x + D) {(x^{2} + 4)}^{2}}{x^{2} + 4}$

Step 1.4

Cancel the common factor of ${(x^{2} + 4)}^{2}$ .

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Step 1.4.1

Cancel the common factor.

$\frac{(x^{2} + 6 x + 1) {(x^{2} + 4)}^{2}}{{(x^{2} + 4)}^{2}} = \frac{(A x + B) {(x^{2} + 4)}^{2}}{{(x^{2} + 4)}^{2}} + \frac{(C x + D) {(x^{2} + 4)}^{2}}{x^{2} + 4}$

Step 1.4.2

Divide $x^{2} + 6 x + 1$ by $1$ .

$x^{2} + 6 x + 1 = \frac{(A x + B) {(x^{2} + 4)}^{2}}{{(x^{2} + 4)}^{2}} + \frac{(C x + D) {(x^{2} + 4)}^{2}}{x^{2} + 4}$

Step 1.5

Simplify each term.

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Step 1.5.1

Cancel the common factor of ${(x^{2} + 4)}^{2}$ .

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Step 1.5.1.1

Cancel the common factor.

$x^{2} + 6 x + 1 = \frac{(A x + B) {(x^{2} + 4)}^{2}}{{(x^{2} + 4)}^{2}} + \frac{(C x + D) {(x^{2} + 4)}^{2}}{x^{2} + 4}$

Step 1.5.1.2

Divide $A x + B$ by $1$ .

$x^{2} + 6 x + 1 = A x + B + \frac{(C x + D) {(x^{2} + 4)}^{2}}{x^{2} + 4}$

Step 1.5.2

Cancel the common factor of ${(x^{2} + 4)}^{2}$ and $x^{2} + 4$ .

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Step 1.5.2.1

Factor $x^{2} + 4$ out of $(C x + D) {(x^{2} + 4)}^{2}$ .

$x^{2} + 6 x + 1 = A x + B + \frac{(x^{2} + 4) ((C x + D) (x^{2} + 4))}{x^{2} + 4}$

Step 1.5.2.2

Cancel the common factors.

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Step 1.5.2.2.1

Multiply by $1$ .

$x^{2} + 6 x + 1 = A x + B + \frac{(x^{2} + 4) ((C x + D) (x^{2} + 4))}{(x^{2} + 4) \cdot 1}$

Step 1.5.2.2.2

Cancel the common factor.

$x^{2} + 6 x + 1 = A x + B + \frac{(x^{2} + 4) ((C x + D) (x^{2} + 4))}{(x^{2} + 4) \cdot 1}$

Step 1.5.2.2.3

Rewrite the expression.

$x^{2} + 6 x + 1 = A x + B + \frac{(C x + D) (x^{2} + 4)}{1}$

Step 1.5.2.2.4

Divide $(C x + D) (x^{2} + 4)$ by $1$ .

$x^{2} + 6 x + 1 = A x + B + (C x + D) (x^{2} + 4)$

Step 1.5.3

Expand $(C x + D) (x^{2} + 4)$ using the FOIL Method.

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Step 1.5.3.1

Apply the distributive property.

$x^{2} + 6 x + 1 = A x + B + C x (x^{2} + 4) + D (x^{2} + 4)$

Step 1.5.3.2

Apply the distributive property.

$x^{2} + 6 x + 1 = A x + B + C x \cdot x^{2} + C x \cdot 4 + D (x^{2} + 4)$

Step 1.5.3.3

Apply the distributive property.

$x^{2} + 6 x + 1 = A x + B + C x \cdot x^{2} + C x \cdot 4 + D x^{2} + D \cdot 4$

Step 1.5.4

Simplify each term.

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Step 1.5.4.1

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

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Step 1.5.4.1.1

Move $x^{2}$ .

$x^{2} + 6 x + 1 = A x + B + C (x^{2} x) + C x \cdot 4 + D x^{2} + D \cdot 4$

Step 1.5.4.1.2

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

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Step 1.5.4.1.2.1

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

$x^{2} + 6 x + 1 = A x + B + C (x^{2} x) + C x \cdot 4 + D x^{2} + D \cdot 4$

Step 1.5.4.1.2.2

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

$x^{2} + 6 x + 1 = A x + B + C x^{2 + 1} + C x \cdot 4 + D x^{2} + D \cdot 4$

Step 1.5.4.1.3

Add $2$ and $1$ .

$x^{2} + 6 x + 1 = A x + B + C x^{3} + C x \cdot 4 + D x^{2} + D \cdot 4$

Step 1.5.4.2

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

$x^{2} + 6 x + 1 = A x + B + C x^{3} + 4 \cdot (C x) + D x^{2} + D \cdot 4$

Step 1.5.4.3

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

$x^{2} + 6 x + 1 = A x + B + C x^{3} + 4 C x + D x^{2} + 4 D$

Step 1.6

Simplify the expression.

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Step 1.6.1

Move $B$ .

$x^{2} + 6 x + 1 = A x + C x^{3} + 4 C x + D x^{2} + B + 4 D$

Step 1.6.2

Move $4 C x$ .

$x^{2} + 6 x + 1 = A x + C x^{3} + D x^{2} + 4 C x + B + 4 D$

Step 1.6.3

Move $A x$ .

$x^{2} + 6 x + 1 = C x^{3} + D x^{2} + A x + 4 C x + B + 4 D$

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

$0 = C$

Step 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.

$1 = D$

Step 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.

$6 = A + 4 C$

Step 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.

$1 = B + 4 D$

Step 2.5

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

$0 = C$

$1 = D$

$6 = A + 4 C$

$1 = B + 4 D$

$0 = C$

$1 = D$

$6 = A + 4 C$

$1 = B + 4 D$

Step 3

Solve the system of equations.

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Step 3.1

Rewrite the equation as $C = 0$ .

$C = 0$

$1 = D$

$6 = A + 4 C$

$1 = B + 4 D$

Step 3.2

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

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Step 3.2.1

Rewrite the equation as $D = 1$ .

$D = 1$

$C = 0$

$6 = A + 4 C$

$1 = B + 4 D$

Step 3.2.2

Replace all occurrences of $C$ in $6 = A + 4 C$ with $0$ .

$6 = A + 4 (0)$

$D = 1$

$C = 0$

$1 = B + 4 D$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify $A + 4 (0)$ .

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Step 3.2.3.1.1

Multiply $4$ by $0$ .

$6 = A + 0$

$D = 1$

$C = 0$

$1 = B + 4 D$

Step 3.2.3.1.2

Add $A$ and $0$ .

$6 = A$

$D = 1$

$C = 0$

$1 = B + 4 D$

$6 = A$

$D = 1$

$C = 0$

$1 = B + 4 D$

$6 = A$

$D = 1$

$C = 0$

$1 = B + 4 D$

$6 = A$

$D = 1$

$C = 0$

$1 = B + 4 D$

Step 3.3

Replace all occurrences of $D$ with $1$ in each equation.

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Step 3.3.1

Rewrite the equation as $A = 6$ .

$A = 6$

$D = 1$

$C = 0$

$1 = B + 4 D$

Step 3.3.2

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

$1 = B + 4 (1)$

$A = 6$

$D = 1$

$C = 0$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Multiply $4$ by $1$ .

$1 = B + 4$

$A = 6$

$D = 1$

$C = 0$

$1 = B + 4$

$A = 6$

$D = 1$

$C = 0$

$1 = B + 4$

$A = 6$

$D = 1$

$C = 0$

Step 3.4

Solve for $B$ in $1 = B + 4$ .

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Step 3.4.1

Rewrite the equation as $B + 4 = 1$ .

$B + 4 = 1$

$A = 6$

$D = 1$

$C = 0$

Step 3.4.2

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

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Step 3.4.2.1

Subtract $4$ from both sides of the equation.

$B = 1 - 4$

$A = 6$

$D = 1$

$C = 0$

Step 3.4.2.2

Subtract $4$ from $1$ .

$B = - 3$

$A = 6$

$D = 1$

$C = 0$

$B = - 3$

$A = 6$

$D = 1$

$C = 0$

$B = - 3$

$A = 6$

$D = 1$

$C = 0$

Step 3.5

Solve the system of equations.

$B = - 3 A = 6 D = 1 C = 0$

Step 3.6

List all of the solutions.

$B = - 3, A = 6, D = 1, C = 0$

Step 4

Replace each of the partial fraction coefficients in $\frac{A x + B}{{(x^{2} + 4)}^{2}} + \frac{C x + D}{x^{2} + 4}$ with the values found for $A$ , $B$ , $C$ , and $D$ .

$\frac{6 x - 3}{{(x^{2} + 4)}^{2}} + \frac{0 x + 1}{x^{2} + 4}$

Step 5

Simplify.

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Step 5.1

Multiply $6$ by $x$ .

$\frac{6 x - 3}{{(x^{2} + 4)}^{2}} + \frac{(0) \cdot x + 1}{x^{2} + 4}$

Step 5.2

Multiply $0$ by $x$ .

$\frac{6 x - 3}{{(x^{2} + 4)}^{2}} + \frac{0 + 1}{x^{2} + 4}$

Step 5.3

Add $0$ and $1$ .

$\frac{6 x - 3}{{(x^{2} + 4)}^{2}} + \frac{1}{x^{2} + 4}$