Calculus Examples

$\frac{1}{(x + 1) (x + 2)}$

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 in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{x + 1}$

Step 1.2

$\frac{A}{x + 1} + \frac{B}{x + 2}$

Step 1.3

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

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

Step 1.4

Cancel the common factor of $x + 1$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Cancel the common factor of $x + 2$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x + 1$ .

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Step 1.6.1.1

Cancel the common factor.

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

Step 1.6.1.2

Divide $(A) (x + 2)$ by $1$ .

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

Step 1.6.2

Apply the distributive property.

$1 = A x + A \cdot 2 + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 1.6.3

Move $2$ to the left of $A$ .

$1 = A x + 2 \cdot A + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 1.6.4

Cancel the common factor of $x + 2$ .

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Step 1.6.4.1

Cancel the common factor.

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

Step 1.6.4.2

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

$1 = A x + 2 A + (B) (x + 1)$

Step 1.6.5

Apply the distributive property.

$1 = A x + 2 A + B x + B \cdot 1$

Step 1.6.6

Multiply $B$ by $1$ .

$1 = A x + 2 A + B x + B$

Step 1.7

Move $2 A$ .

$1 = A x + B x + 2 A + B$

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

$0 = A + B$

Step 2.2

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.

$1 = 2 A + B$

Step 2.3

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

$0 = A + B$

$1 = 2 A + B$

$0 = A + B$

$1 = 2 A + B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $0 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = 0$ .

$A + B = 0$

$1 = 2 A + B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$1 = 2 A + B$

$A = - B$

$1 = 2 A + B$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $A$ in $1 = 2 A + B$ with $- B$ .

$1 = 2 (- B) + B$

$A = - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (- B) + B$ .

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Step 3.2.2.1.1

Multiply $- 1$ by $2$ .

$1 = - 2 B + B$

$A = - B$

Step 3.2.2.1.2

Add $- 2 B$ and $B$ .

$1 = - B$

$A = - B$

$1 = - B$

$A = - B$

$1 = - B$

$A = - B$

$1 = - B$

$A = - B$

Step 3.3

Solve for $B$ in $1 = - B$ .

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Step 3.3.1

Rewrite the equation as $- B = 1$ .

$- B = 1$

$A = - B$

Step 3.3.2

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

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Step 3.3.2.1

Divide each term in $- B = 1$ by $- 1$ .

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

$A = - B$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Dividing two negative values results in a positive value.

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

$A = - B$

Step 3.3.2.2.2

Divide $B$ by $1$ .

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

$A = - B$

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

$A = - B$

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Divide $1$ by $- 1$ .

$B = - 1$

$A = - B$

$B = - 1$

$A = - B$

$B = - 1$

$A = - B$

$B = - 1$

$A = - B$

Step 3.4

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

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Step 3.4.1

Replace all occurrences of $B$ in $A = - B$ with $- 1$ .

$A = - (- 1)$

$B = - 1$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Multiply $- 1$ by $- 1$ .

$A = 1$

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

$B = - 1$

Step 3.5

List all of the solutions.

$A = 1, B = - 1$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x + 1} + \frac{B}{x + 2}$ with the values found for $A$ and $B$ .

$\frac{1}{x + 1} + \frac{- 1}{x + 2}$