Calculus Examples

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

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 the fraction.

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Step 1.1.1.1

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

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

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

Step 1.1.1.1.3

Factor $x$ out of $- 2 x$ .

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

Step 1.1.1.1.4

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

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

Step 1.1.1.1.5

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

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

Step 1.1.1.2

Factor.

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Step 1.1.1.2.1

Factor $x^{2} + x - 2$ using the AC method.

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Step 1.1.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 1.1.1.2.1.2

Write the factored form using these integers.

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

Step 1.1.1.2.2

Remove unnecessary parentheses.

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

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 $B$ .

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

Step 1.1.3

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

Step 1.1.4

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

$\frac{(4 x^{2} - 3 x - 4) (x (x - 1) (x + 2))}{x (x - 1) (x + 2)} = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.5

Cancel the common factor of $x$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(4 x^{2} - 3 x - 4) (x (x - 1) (x + 2))}{x (x - 1) (x + 2)} = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(4 x^{2} - 3 x - 4) ((x - 1) (x + 2))}{(x - 1) (x + 2)} = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.6

Cancel the common factor of $x - 1$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(4 x^{2} - 3 x - 4) ((x - 1) (x + 2))}{(x - 1) (x + 2)} = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.6.2

Rewrite the expression.

$\frac{(4 x^{2} - 3 x - 4) (x + 2)}{x + 2} = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.7

Cancel the common factor of $x + 2$ .

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Step 1.1.7.1

Cancel the common factor.

$\frac{(4 x^{2} - 3 x - 4) (x + 2)}{x + 2} = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.7.2

Divide $4 x^{2} - 3 x - 4$ by $1$ .

$4 x^{2} - 3 x - 4 = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8

Simplify each term.

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Step 1.1.8.1

Cancel the common factor of $x$ .

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Step 1.1.8.1.1

Cancel the common factor.

$4 x^{2} - 3 x - 4 = \frac{A (x (x - 1) (x + 2))}{x} + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.1.2

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

$4 x^{2} - 3 x - 4 = A ((x - 1) (x + 2)) + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.2

Expand $(x - 1) (x + 2)$ using the FOIL Method.

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Step 1.1.8.2.1

Apply the distributive property.

$4 x^{2} - 3 x - 4 = A (x (x + 2) - 1 (x + 2)) + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.2.2

Apply the distributive property.

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

Step 1.1.8.2.3

Apply the distributive property.

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

Step 1.1.8.3

Simplify and combine like terms.

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Step 1.1.8.3.1

Simplify each term.

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Step 1.1.8.3.1.1

Multiply $x$ by $x$ .

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

Step 1.1.8.3.1.2

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

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

Step 1.1.8.3.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 1.1.8.3.1.4

Multiply $- 1$ by $2$ .

$4 x^{2} - 3 x - 4 = A (x^{2} + 2 x - x - 2) + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.3.2

Subtract $x$ from $2 x$ .

$4 x^{2} - 3 x - 4 = A (x^{2} + x - 2) + \frac{(B) (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.4

Apply the distributive property.

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

Step 1.1.8.5

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

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

Step 1.1.8.6

Cancel the common factor of $x - 1$ .

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Step 1.1.8.6.1

Cancel the common factor.

$4 x^{2} - 3 x - 4 = A x^{2} + A x - 2 A + \frac{B (x (x - 1) (x + 2))}{x - 1} + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.6.2

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

$4 x^{2} - 3 x - 4 = A x^{2} + A x - 2 A + (B) (x (x + 2)) + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.7

Apply the distributive property.

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

Step 1.1.8.8

Multiply $x$ by $x$ .

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

Step 1.1.8.9

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

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

Step 1.1.8.10

Apply the distributive property.

$4 x^{2} - 3 x - 4 = A x^{2} + A x - 2 A + B x^{2} + B (2 x) + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.11

Rewrite using the commutative property of multiplication.

$4 x^{2} - 3 x - 4 = A x^{2} + A x - 2 A + B x^{2} + 2 B x + \frac{(C) (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.12

Cancel the common factor of $x + 2$ .

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Step 1.1.8.12.1

Cancel the common factor.

$4 x^{2} - 3 x - 4 = A x^{2} + A x - 2 A + B x^{2} + 2 B x + \frac{C (x (x - 1) (x + 2))}{x + 2}$

Step 1.1.8.12.2

Divide $(C) (x (x - 1))$ by $1$ .

$4 x^{2} - 3 x - 4 = A x^{2} + A x - 2 A + B x^{2} + 2 B x + (C) (x (x - 1))$

Step 1.1.8.13

Apply the distributive property.

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

Step 1.1.8.14

Multiply $x$ by $x$ .

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

Step 1.1.8.15

Move $- 1$ to the left of $x$ .

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

Step 1.1.8.16

Rewrite $- 1 x$ as $- x$ .

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

Step 1.1.8.17

Apply the distributive property.

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

Step 1.1.8.18

Rewrite using the commutative property of multiplication.

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

Step 1.1.9

Simplify the expression.

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Step 1.1.9.1

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

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

Step 1.1.9.2

Move $B$ .

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

Step 1.1.9.3

Move $- 2 A$ .

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

Step 1.1.9.4

Move $2 x B$ .

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

Step 1.1.9.5

Move $A x$ .

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

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$4 = A + B + C$

Step 1.2.2

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.

$- 3 = A + 2 B - 1 C$

Step 1.2.3

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.

$- 4 = - 2 A$

Step 1.2.4

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

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$- 4 = - 2 A$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$- 4 = - 2 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $- 4 = - 2 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 2 A = - 4$ .

$- 2 A = - 4$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

Step 1.3.1.2

Divide each term in $- 2 A = - 4$ by $- 2$ and simplify.

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Step 1.3.1.2.1

Divide each term in $- 2 A = - 4$ by $- 2$ .

$\frac{- 2 A}{- 2} = \frac{- 4}{- 2}$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

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 $- 2$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 2 A}{- 2} = \frac{- 4}{- 2}$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 4}{- 2}$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$A = \frac{- 4}{- 2}$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$A = \frac{- 4}{- 2}$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $- 4$ by $- 2$ .

$A = 2$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$A = 2$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$A = 2$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

$A = 2$

$4 = A + B + C$

$- 3 = A + 2 B - 1 C$

Step 1.3.2

Replace all occurrences of $A$ with $2$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $4 = A + B + C$ with $2$ .

$4 = (2) + B + C$

$A = 2$

$- 3 = A + 2 B - 1 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$4 = 2 + B + C$

$A = 2$

$- 3 = A + 2 B - 1 C$

$4 = 2 + B + C$

$A = 2$

$- 3 = A + 2 B - 1 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $- 3 = A + 2 B - 1 C$ with $2$ .

$- 3 = (2) + 2 B - 1 C$

$4 = 2 + B + C$

$A = 2$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Rewrite $- 1 C$ as $- C$ .

$- 3 = 2 + 2 B - C$

$4 = 2 + B + C$

$A = 2$

$- 3 = 2 + 2 B - C$

$4 = 2 + B + C$

$A = 2$

$- 3 = 2 + 2 B - C$

$4 = 2 + B + C$

$A = 2$

Step 1.3.3

Solve for $B$ in $4 = 2 + B + C$ .

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Step 1.3.3.1

Rewrite the equation as $2 + B + C = 4$ .

$2 + B + C = 4$

$- 3 = 2 + 2 B - C$

$A = 2$

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 $2$ from both sides of the equation.

$B + C = 4 - 2$

$- 3 = 2 + 2 B - C$

$A = 2$

Step 1.3.3.2.2

Subtract $C$ from both sides of the equation.

$B = 4 - 2 - C$

$- 3 = 2 + 2 B - C$

$A = 2$

Step 1.3.3.2.3

Subtract $2$ from $4$ .

$B = 2 - C$

$- 3 = 2 + 2 B - C$

$A = 2$

$B = 2 - C$

$- 3 = 2 + 2 B - C$

$A = 2$

$B = 2 - C$

$- 3 = 2 + 2 B - C$

$A = 2$

Step 1.3.4

Replace all occurrences of $B$ with $2 - C$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $- 3 = 2 + 2 B - C$ with $2 - C$ .

$- 3 = 2 + 2 (2 - C) - C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $2 + 2 (2 - C) - C$ .

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Step 1.3.4.2.1.1

Simplify each term.

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Step 1.3.4.2.1.1.1

Apply the distributive property.

$- 3 = 2 + 2 \cdot 2 + 2 (- C) - C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.1.2

Multiply $2$ by $2$ .

$- 3 = 2 + 4 + 2 (- C) - C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.1.3

Multiply $- 1$ by $2$ .

$- 3 = 2 + 4 - 2 C - C$

$B = 2 - C$

$A = 2$

$- 3 = 2 + 4 - 2 C - C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.2

Simplify by adding terms.

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Step 1.3.4.2.1.2.1

Add $2$ and $4$ .

$- 3 = 6 - 2 C - C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.2.2

Subtract $C$ from $- 2 C$ .

$- 3 = 6 - 3 C$

$B = 2 - C$

$A = 2$

$- 3 = 6 - 3 C$

$B = 2 - C$

$A = 2$

$- 3 = 6 - 3 C$

$B = 2 - C$

$A = 2$

$- 3 = 6 - 3 C$

$B = 2 - C$

$A = 2$

$- 3 = 6 - 3 C$

$B = 2 - C$

$A = 2$

Step 1.3.5

Solve for $C$ in $- 3 = 6 - 3 C$ .

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Step 1.3.5.1

Rewrite the equation as $6 - 3 C = - 3$ .

$6 - 3 C = - 3$

$B = 2 - C$

$A = 2$

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 $6$ from both sides of the equation.

$- 3 C = - 3 - 6$

$B = 2 - C$

$A = 2$

Step 1.3.5.2.2

Subtract $6$ from $- 3$ .

$- 3 C = - 9$

$B = 2 - C$

$A = 2$

$- 3 C = - 9$

$B = 2 - C$

$A = 2$

Step 1.3.5.3

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

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Step 1.3.5.3.1

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

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

$B = 2 - C$

$A = 2$

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.

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

$B = 2 - C$

$A = 2$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

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

$B = 2 - C$

$A = 2$

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

$B = 2 - C$

$A = 2$

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

$B = 2 - C$

$A = 2$

Step 1.3.5.3.3

Simplify the right side.

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Step 1.3.5.3.3.1

Divide $- 9$ by $- 3$ .

$C = 3$

$B = 2 - C$

$A = 2$

$C = 3$

$B = 2 - C$

$A = 2$

$C = 3$

$B = 2 - C$

$A = 2$

$C = 3$

$B = 2 - C$

$A = 2$

Step 1.3.6

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

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Step 1.3.6.1

Replace all occurrences of $C$ in $B = 2 - C$ with $3$ .

$B = 2 - (3)$

$C = 3$

$A = 2$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $2 - (3)$ .

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Step 1.3.6.2.1.1

Multiply $- 1$ by $3$ .

$B = 2 - 3$

$C = 3$

$A = 2$

Step 1.3.6.2.1.2

Subtract $3$ from $2$ .

$B = - 1$

$C = 3$

$A = 2$

$B = - 1$

$C = 3$

$A = 2$

$B = - 1$

$C = 3$

$A = 2$

$B = - 1$

$C = 3$

$A = 2$

Step 1.3.7

List all of the solutions.

$B = - 1, C = 3, A = 2$

Step 1.4

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

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

Step 1.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 4

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

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

Step 5

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

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

Step 6

Let $u_{1} = x - 1$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 6.1

Let $u_{1} = x - 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 6.1.1

Differentiate $x - 1$ .

$\frac{d}{d x} [x - 1]$

Step 6.1.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 6.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- 1]$

Step 6.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

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

Step 7

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

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

Step 8

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

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

Step 9

Let $u_{2} = x + 2$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1

Let $u_{2} = x + 2$ . Find $\frac{d u_{2}}{d x}$ .

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Step 9.1.1

Differentiate $x + 2$ .

$\frac{d}{d x} [x + 2]$

Step 9.1.2

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 9.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [2]$

Step 9.1.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

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

Step 10

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$2 (\ln (| x |) + C) - (\ln (| u_{1} |) + C) + 3 (\ln (| u_{2} |) + C)$

Step 11

Simplify.

$2 \ln (| x |) - \ln (| u_{1} |) + 3 \ln (| u_{2} |) + C$

Step 12

Substitute back in for each integration substitution variable.

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Step 12.1

Replace all occurrences of $u_{1}$ with $x - 1$ .

$2 \ln (| x |) - \ln (| x - 1 |) + 3 \ln (| u_{2} |) + C$

Step 12.2

Replace all occurrences of $u_{2}$ with $x + 2$ .

$2 \ln (| x |) - \ln (| x - 1 |) + 3 \ln (| x + 2 |) + C$