Calculus Examples

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

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

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

$\frac{5 x + 3}{x \cdot x^{2} + x (- 2 x) - 3 x}$

Step 1.1.1.1.3

Factor $x$ out of $- 3 x$ .

$\frac{5 x + 3}{x \cdot x^{2} + x (- 2 x) + x \cdot - 3}$

Step 1.1.1.1.4

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

$\frac{5 x + 3}{x (x^{2} - 2 x) + x \cdot - 3}$

Step 1.1.1.1.5

Factor $x$ out of $x (x^{2} - 2 x) + x \cdot - 3$ .

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

Step 1.1.1.2

Factor.

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Step 1.1.1.2.1

Factor $x^{2} - 2 x - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 1.1.1.2.1.2

Write the factored form using these integers.

$\frac{5 x + 3}{x ((x - 3) (x + 1))}$

Step 1.1.1.2.2

Remove unnecessary parentheses.

$\frac{5 x + 3}{x (x - 3) (x + 1)}$

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 $B$ .

$\frac{A}{x} + \frac{B}{x - 3}$

Step 1.1.3

$\frac{A}{x} + \frac{B}{x - 3} + \frac{C}{x + 1}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x - 3) (x + 1)$ .

$\frac{(5 x + 3) (x (x - 3) (x + 1))}{x (x - 3) (x + 1)} = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.5

Cancel the common factor of $x$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(5 x + 3) (x (x - 3) (x + 1))}{x (x - 3) (x + 1)} = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(5 x + 3) ((x - 3) (x + 1))}{(x - 3) (x + 1)} = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.6

Cancel the common factor of $x - 3$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(5 x + 3) ((x - 3) (x + 1))}{(x - 3) (x + 1)} = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.6.2

Rewrite the expression.

$\frac{(5 x + 3) (x + 1)}{x + 1} = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.7

Cancel the common factor of $x + 1$ .

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Step 1.1.7.1

Cancel the common factor.

$\frac{(5 x + 3) (x + 1)}{x + 1} = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.7.2

Divide $5 x + 3$ by $1$ .

$5 x + 3 = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

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.

$5 x + 3 = \frac{A (x (x - 3) (x + 1))}{x} + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.1.2

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

$5 x + 3 = A ((x - 3) (x + 1)) + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.2

Expand $(x - 3) (x + 1)$ using the FOIL Method.

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Step 1.1.8.2.1

Apply the distributive property.

$5 x + 3 = A (x (x + 1) - 3 (x + 1)) + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.2.2

Apply the distributive property.

$5 x + 3 = A (x \cdot x + x \cdot 1 - 3 (x + 1)) + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.2.3

Apply the distributive property.

$5 x + 3 = A (x \cdot x + x \cdot 1 - 3 x - 3 \cdot 1) + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

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 $x$ .

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

Step 1.1.8.3.1.2

Multiply $x$ by $1$ .

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

Step 1.1.8.3.1.3

Multiply $- 3$ by $1$ .

$5 x + 3 = A (x^{2} + x - 3 x - 3) + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.3.2

Subtract $3 x$ from $x$ .

$5 x + 3 = A (x^{2} - 2 x - 3) + \frac{(B) (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.4

Apply the distributive property.

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

Step 1.1.8.5

Simplify.

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Step 1.1.8.5.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.5.2

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

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

Step 1.1.8.6

Cancel the common factor of $x - 3$ .

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Step 1.1.8.6.1

Cancel the common factor.

$5 x + 3 = A x^{2} - 2 A x - 3 A + \frac{B (x (x - 3) (x + 1))}{x - 3} + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.6.2

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

$5 x + 3 = A x^{2} - 2 A x - 3 A + (B) (x (x + 1)) + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.7

Apply the distributive property.

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

Step 1.1.8.8

Multiply $x$ by $x$ .

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

Step 1.1.8.9

Multiply $x$ by $1$ .

$5 x + 3 = A x^{2} - 2 A x - 3 A + B (x^{2} + x) + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.10

Apply the distributive property.

$5 x + 3 = A x^{2} - 2 A x - 3 A + B x^{2} + B x + \frac{(C) (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.11

Cancel the common factor of $x + 1$ .

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Step 1.1.8.11.1

Cancel the common factor.

$5 x + 3 = A x^{2} - 2 A x - 3 A + B x^{2} + B x + \frac{C (x (x - 3) (x + 1))}{x + 1}$

Step 1.1.8.11.2

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

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

Step 1.1.8.12

Apply the distributive property.

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

Step 1.1.8.13

Multiply $x$ by $x$ .

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

Step 1.1.8.14

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

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

Step 1.1.8.15

Apply the distributive property.

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

Step 1.1.8.16

Rewrite using the commutative property of multiplication.

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

Step 1.1.9

Simplify the expression.

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Step 1.1.9.1

Move $A$ .

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

Step 1.1.9.2

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

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

Step 1.1.9.3

Reorder $B$ and $x$ .

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

Step 1.1.9.4

Move $- 3 A$ .

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

Step 1.1.9.5

Move $B x$ .

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

Step 1.1.9.6

Move $- 2 x A$ .

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

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

$0 = A + B + C$

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.

$5 = - 2 A + B - 3 C$

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.

$3 = - 3 A$

Step 1.2.4

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

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

$3 = - 3 A$

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

$3 = - 3 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

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

$- 3 A = 3$

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

Step 1.3.1.2

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

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Step 1.3.1.2.1

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

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

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

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 $- 3$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

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

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

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

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

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

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

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

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $3$ by $- 3$ .

$A = - 1$

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

$A = - 1$

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

$A = - 1$

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

$A = - 1$

$0 = A + B + C$

$5 = - 2 A + B - 3 C$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $0 = A + B + C$ with $- 1$ .

$0 = (- 1) + B + C$

$A = - 1$

$5 = - 2 A + B - 3 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$0 = - 1 + B + C$

$A = - 1$

$5 = - 2 A + B - 3 C$

$0 = - 1 + B + C$

$A = - 1$

$5 = - 2 A + B - 3 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $5 = - 2 A + B - 3 C$ with $- 1$ .

$5 = - 2 \cdot - 1 + B - 3 C$

$0 = - 1 + B + C$

$A = - 1$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Multiply $- 2$ by $- 1$ .

$5 = 2 + B - 3 C$

$0 = - 1 + B + C$

$A = - 1$

$5 = 2 + B - 3 C$

$0 = - 1 + B + C$

$A = - 1$

$5 = 2 + B - 3 C$

$0 = - 1 + B + C$

$A = - 1$

Step 1.3.3

Solve for $B$ in $5 = 2 + B - 3 C$ .

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Step 1.3.3.1

Rewrite the equation as $2 + B - 3 C = 5$ .

$2 + B - 3 C = 5$

$0 = - 1 + B + C$

$A = - 1$

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.

$B - 3 C = 5 - 2$

$0 = - 1 + B + C$

$A = - 1$

Step 1.3.3.2.2

Add $3 C$ to both sides of the equation.

$B = 5 - 2 + 3 C$

$0 = - 1 + B + C$

$A = - 1$

Step 1.3.3.2.3

Subtract $2$ from $5$ .

$B = 3 + 3 C$

$0 = - 1 + B + C$

$A = - 1$

$B = 3 + 3 C$

$0 = - 1 + B + C$

$A = - 1$

$B = 3 + 3 C$

$0 = - 1 + B + C$

$A = - 1$

Step 1.3.4

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

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Step 1.3.4.1

Replace all occurrences of $B$ in $0 = - 1 + B + C$ with $3 + 3 C$ .

$0 = - 1 + 3 + 3 C + C$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.4.2

Simplify $0 = - 1 + 3 + 3 C + C$ .

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Step 1.3.4.2.1

Simplify the left side.

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Step 1.3.4.2.1.1

Remove parentheses.

$0 = - 1 + 3 + 3 C + C$

$B = 3 + 3 C$

$A = - 1$

$0 = - 1 + 3 + 3 C + C$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.4.2.2

Simplify the right side.

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Step 1.3.4.2.2.1

Simplify $- 1 + 3 + 3 C + C$ .

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Step 1.3.4.2.2.1.1

Add $- 1$ and $3$ .

$0 = 2 + 3 C + C$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.4.2.2.1.2

Add $3 C$ and $C$ .

$0 = 2 + 4 C$

$B = 3 + 3 C$

$A = - 1$

$0 = 2 + 4 C$

$B = 3 + 3 C$

$A = - 1$

$0 = 2 + 4 C$

$B = 3 + 3 C$

$A = - 1$

$0 = 2 + 4 C$

$B = 3 + 3 C$

$A = - 1$

$0 = 2 + 4 C$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5

Solve for $C$ in $0 = 2 + 4 C$ .

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Step 1.3.5.1

Rewrite the equation as $2 + 4 C = 0$ .

$2 + 4 C = 0$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.2

Subtract $2$ from both sides of the equation.

$4 C = - 2$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3

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

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Step 1.3.5.3.1

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

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

$B = 3 + 3 C$

$A = - 1$

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 $4$ .

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Step 1.3.5.3.2.1.1

Cancel the common factor.

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

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

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

$B = 3 + 3 C$

$A = - 1$

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

$B = 3 + 3 C$

$A = - 1$

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

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3.3

Simplify the right side.

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Step 1.3.5.3.3.1

Cancel the common factor of $- 2$ and $4$ .

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Step 1.3.5.3.3.1.1

Factor $2$ out of $- 2$ .

$C = \frac{2 (- 1)}{4}$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3.3.1.2

Cancel the common factors.

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Step 1.3.5.3.3.1.2.1

Factor $2$ out of $4$ .

$C = \frac{2 \cdot - 1}{2 \cdot 2}$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3.3.1.2.2

Cancel the common factor.

$C = \frac{2 \cdot - 1}{2 \cdot 2}$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3.3.1.2.3

Rewrite the expression.

$C = \frac{- 1}{2}$

$B = 3 + 3 C$

$A = - 1$

$C = \frac{- 1}{2}$

$B = 3 + 3 C$

$A = - 1$

$C = \frac{- 1}{2}$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.5.3.3.2

Move the negative in front of the fraction.

$C = - \frac{1}{2}$

$B = 3 + 3 C$

$A = - 1$

$C = - \frac{1}{2}$

$B = 3 + 3 C$

$A = - 1$

$C = - \frac{1}{2}$

$B = 3 + 3 C$

$A = - 1$

$C = - \frac{1}{2}$

$B = 3 + 3 C$

$A = - 1$

Step 1.3.6

Replace all occurrences of $C$ with $- \frac{1}{2}$ in each equation.

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Step 1.3.6.1

Replace all occurrences of $C$ in $B = 3 + 3 C$ with $- \frac{1}{2}$ .

$B = 3 + 3 (- \frac{1}{2})$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $3 + 3 (- \frac{1}{2})$ .

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Step 1.3.6.2.1.1

Simplify each term.

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Step 1.3.6.2.1.1.1

Multiply $3 (- \frac{1}{2})$ .

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Step 1.3.6.2.1.1.1.1

Multiply $- 1$ by $3$ .

$B = 3 - 3 (\frac{1}{2})$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.1.1.2

Combine $- 3$ and $\frac{1}{2}$ .

$B = 3 + \frac{- 3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

$B = 3 + \frac{- 3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.1.2

Move the negative in front of the fraction.

$B = 3 - \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

$B = 3 - \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$B = 3 \cdot \frac{2}{2} - \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.3

Combine $3$ and $\frac{2}{2}$ .

$B = \frac{3 \cdot 2}{2} - \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.4

Combine the numerators over the common denominator.

$B = \frac{3 \cdot 2 - 3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.5

Simplify the numerator.

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Step 1.3.6.2.1.5.1

Multiply $3$ by $2$ .

$B = \frac{6 - 3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.6.2.1.5.2

Subtract $3$ from $6$ .

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = - 1$

Step 1.3.7

List all of the solutions.

$B = \frac{3}{2}, C = - \frac{1}{2}, A = - 1$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x - 3} + \frac{C}{x + 1}$ with the values found for $A$ , $B$ , and $C$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.2

Multiply $\frac{3}{2}$ by $\frac{1}{x - 3}$ .

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

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.4

Multiply $\frac{1}{x + 1}$ by $\frac{1}{2}$ .

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

Step 1.5.5

Move $2$ to the left of $x + 1$ .

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

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

Step 5

Since $\frac{3}{2}$ is constant with respect to $x$ , move $\frac{3}{2}$ out of the integral.

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

Step 6

Let $u_{1} = x - 3$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 6.1

Let $u_{1} = x - 3$ . Find $\frac{d u_{1}}{d x}$ .

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Step 6.1.1

Differentiate $x - 3$ .

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

Step 6.1.2

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 6.1.3

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

$1 + \frac{d}{d x} [- 3]$

Step 6.1.4

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ 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 $d u_{1}$ .

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

Step 7

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

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

Step 8

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

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

Step 9

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

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

Step 10

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 10.1

Let $u_{2} = x + 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 10.1.1

Differentiate $x + 1$ .

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

Step 10.1.2

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 10.1.3

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 10.1.4

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

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

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

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

Step 11

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

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

Step 12

Simplify.

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

Step 13

Substitute back in for each integration substitution variable.

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Step 13.1

Replace all occurrences of $u_{1}$ with $x - 3$ .

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

Step 13.2

Replace all occurrences of $u_{2}$ with $x + 1$ .

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