Calculus Examples

$\int \frac{3 x^{2} - 3 x + 6}{4 x^{3} - 6 x^{2} + 24 x} d x$

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 $3$ out of $3 x^{2} - 3 x + 6$ .

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Step 1.1.1.1.1

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

$\frac{3 (x^{2}) - 3 x + 6}{4 x^{3} - 6 x^{2} + 24 x}$

Step 1.1.1.1.2

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

$\frac{3 (x^{2}) + 3 (- x) + 6}{4 x^{3} - 6 x^{2} + 24 x}$

Step 1.1.1.1.3

Factor $3$ out of $6$ .

$\frac{3 (x^{2}) + 3 (- x) + 3 (2)}{4 x^{3} - 6 x^{2} + 24 x}$

Step 1.1.1.1.4

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

$\frac{3 (x^{2} - x) + 3 (2)}{4 x^{3} - 6 x^{2} + 24 x}$

Step 1.1.1.1.5

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

$\frac{3 (x^{2} - x + 2)}{4 x^{3} - 6 x^{2} + 24 x}$

Step 1.1.1.2

Factor $2 x$ out of $4 x^{3} - 6 x^{2} + 24 x$ .

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Step 1.1.1.2.1

Factor $2 x$ out of $4 x^{3}$ .

$\frac{3 (x^{2} - x + 2)}{2 x (2 x^{2}) - 6 x^{2} + 24 x}$

Step 1.1.1.2.2

Factor $2 x$ out of $- 6 x^{2}$ .

$\frac{3 (x^{2} - x + 2)}{2 x (2 x^{2}) + 2 x (- 3 x) + 24 x}$

Step 1.1.1.2.3

Factor $2 x$ out of $24 x$ .

$\frac{3 (x^{2} - x + 2)}{2 x (2 x^{2}) + 2 x (- 3 x) + 2 x (12)}$

Step 1.1.1.2.4

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

$\frac{3 (x^{2} - x + 2)}{2 x (2 x^{2} - 3 x) + 2 x (12)}$

Step 1.1.1.2.5

Factor $2 x$ out of $2 x (2 x^{2} - 3 x) + 2 x (12)$ .

$\frac{3 (x^{2} - x + 2)}{2 x (2 x^{2} - 3 x + 12)}$

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 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.

$\frac{A}{x} + \frac{B x + C}{2 x^{2} - 3 x + 12}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $2 x (2 x^{2} - 3 x + 12)$ .

$\frac{3 (x^{2} - x + 2) (2 x (2 x^{2} - 3 x + 12))}{2 x (2 x^{2} - 3 x + 12)} = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.4

Cancel the common factor of $2$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{3 (x^{2} - x + 2) (2 x (2 x^{2} - 3 x + 12))}{2 x (2 x^{2} - 3 x + 12)} = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.4.2

Rewrite the expression.

$\frac{3 (x^{2} - x + 2) (x (2 x^{2} - 3 x + 12))}{x (2 x^{2} - 3 x + 12)} = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.5

Cancel the common factor of $x$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{3 (x^{2} - x + 2) (x (2 x^{2} - 3 x + 12))}{x (2 x^{2} - 3 x + 12)} = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.5.2

Rewrite the expression.

$\frac{3 (x^{2} - x + 2) (2 x^{2} - 3 x + 12)}{2 x^{2} - 3 x + 12} = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.6

Cancel the common factor of $2 x^{2} - 3 x + 12$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{3 (x^{2} - x + 2) (2 x^{2} - 3 x + 12)}{2 x^{2} - 3 x + 12} = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.6.2

Divide $3 (x^{2} - x + 2)$ by $1$ .

$3 (x^{2} - x + 2) = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.7

Apply the distributive property.

$3 x^{2} + 3 (- x) + 3 \cdot 2 = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.8

Simplify.

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Step 1.1.8.1

Multiply $- 1$ by $3$ .

$3 x^{2} - 3 x + 3 \cdot 2 = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.8.2

Multiply $3$ by $2$ .

$3 x^{2} - 3 x + 6 = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9

Simplify each term.

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Step 1.1.9.1

Cancel the common factor of $x$ .

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Step 1.1.9.1.1

Cancel the common factor.

$3 x^{2} - 3 x + 6 = \frac{A (2 x (2 x^{2} - 3 x + 12))}{x} + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.1.2

Divide $A (2 (2 x^{2} - 3 x + 12))$ by $1$ .

$3 x^{2} - 3 x + 6 = A (2 (2 x^{2} - 3 x + 12)) + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.2

Rewrite using the commutative property of multiplication.

$3 x^{2} - 3 x + 6 = 2 A (2 x^{2} - 3 x + 12) + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.3

Apply the distributive property.

$3 x^{2} - 3 x + 6 = 2 A (2 x^{2}) + 2 A (- 3 x) + 2 A \cdot 12 + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.4

Simplify.

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Step 1.1.9.4.1

Rewrite using the commutative property of multiplication.

$3 x^{2} - 3 x + 6 = 2 \cdot (2 A x^{2}) + 2 A (- 3 x) + 2 A \cdot 12 + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.4.2

Rewrite using the commutative property of multiplication.

$3 x^{2} - 3 x + 6 = 2 \cdot (2 A x^{2}) + 2 \cdot (- 3 A x) + 2 A \cdot 12 + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.4.3

Multiply $12$ by $2$ .

$3 x^{2} - 3 x + 6 = 2 \cdot (2 A x^{2}) + 2 \cdot (- 3 A x) + 24 A + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.5

Simplify each term.

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Step 1.1.9.5.1

Multiply $2$ by $2$ .

$3 x^{2} - 3 x + 6 = 4 A x^{2} + 2 \cdot (- 3 A x) + 24 A + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.5.2

Multiply $2$ by $- 3$ .

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.6

Cancel the common factor of $2 x^{2} - 3 x + 12$ .

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Step 1.1.9.6.1

Cancel the common factor.

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + \frac{(B x + C) (2 x (2 x^{2} - 3 x + 12))}{2 x^{2} - 3 x + 12}$

Step 1.1.9.6.2

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

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + (B x + C) (2 x)$

Step 1.1.9.7

Apply the distributive property.

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + B x (2 x) + C (2 x)$

Step 1.1.9.8

Rewrite using the commutative property of multiplication.

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + 2 B x \cdot x + C (2 x)$

Step 1.1.9.9

Rewrite using the commutative property of multiplication.

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + 2 B x \cdot x + 2 C x$

Step 1.1.9.10

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

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Step 1.1.9.10.1

Move $x$ .

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + 2 B (x \cdot x) + 2 C x$

Step 1.1.9.10.2

Multiply $x$ by $x$ .

$3 x^{2} - 3 x + 6 = 4 A x^{2} - 6 A x + 24 A + 2 B x^{2} + 2 C x$

Step 1.1.10

Reorder.

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Step 1.1.10.1

Move $A$ .

$3 x^{2} - 3 x + 6 = 4 x^{2} A - 6 A x + 24 A + 2 B x^{2} + 2 C x$

Step 1.1.10.2

Move $A$ .

$3 x^{2} - 3 x + 6 = 4 x^{2} A - 6 x A + 24 A + 2 B x^{2} + 2 C x$

Step 1.1.10.3

Move $B$ .

$3 x^{2} - 3 x + 6 = 4 x^{2} A - 6 x A + 24 A + 2 x^{2} B + 2 C x$

Step 1.1.10.4

Move $24 A$ .

$3 x^{2} - 3 x + 6 = 4 x^{2} A - 6 x A + 2 x^{2} B + 2 C x + 24 A$

Step 1.1.10.5

Move $- 6 x A$ .

$3 x^{2} - 3 x + 6 = 4 x^{2} A + 2 x^{2} B - 6 x A + 2 C x + 24 A$

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.

$3 = 4 A + 2 B$

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.

$- 3 = - 6 A + 2 C$

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.

$6 = 24 A$

Step 1.2.4

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

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$6 = 24 A$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$6 = 24 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $6 = 24 A$ .

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Step 1.3.1.1

Rewrite the equation as $24 A = 6$ .

$24 A = 6$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.1.2

Divide each term in $24 A = 6$ by $24$ and simplify.

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Step 1.3.1.2.1

Divide each term in $24 A = 6$ by $24$ .

$\frac{24 A}{24} = \frac{6}{24}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

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 $24$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

$\frac{24 A}{24} = \frac{6}{24}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{6}{24}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{6}{24}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{6}{24}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Cancel the common factor of $6$ and $24$ .

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Step 1.3.1.2.3.1.1

Factor $6$ out of $6$ .

$A = \frac{6 (1)}{24}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.1.2.3.1.2

Cancel the common factors.

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Step 1.3.1.2.3.1.2.1

Factor $6$ out of $24$ .

$A = \frac{6 \cdot 1}{6 \cdot 4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.1.2.3.1.2.2

Cancel the common factor.

$A = \frac{6 \cdot 1}{6 \cdot 4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.1.2.3.1.2.3

Rewrite the expression.

$A = \frac{1}{4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{1}{4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{1}{4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{1}{4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{1}{4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

$A = \frac{1}{4}$

$3 = 4 A + 2 B$

$- 3 = - 6 A + 2 C$

Step 1.3.2

Replace all occurrences of $A$ with $\frac{1}{4}$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $3 = 4 A + 2 B$ with $\frac{1}{4}$ .

$3 = 4 (\frac{1}{4}) + 2 B$

$A = \frac{1}{4}$

$- 3 = - 6 A + 2 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Cancel the common factor of $4$ .

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Step 1.3.2.2.1.1

Cancel the common factor.

$3 = 4 (\frac{1}{4}) + 2 B$

$A = \frac{1}{4}$

$- 3 = - 6 A + 2 C$

Step 1.3.2.2.1.2

Rewrite the expression.

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$- 3 = - 6 A + 2 C$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$- 3 = - 6 A + 2 C$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$- 3 = - 6 A + 2 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $- 3 = - 6 A + 2 C$ with $\frac{1}{4}$ .

$- 3 = - 6 (\frac{1}{4}) + 2 C$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify each term.

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Step 1.3.2.4.1.1

Cancel the common factor of $2$ .

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Step 1.3.2.4.1.1.1

Factor $2$ out of $- 6$ .

$- 3 = 2 (- 3) (\frac{1}{4}) + 2 C$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.2.4.1.1.2

Factor $2$ out of $4$ .

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.2.4.1.1.3

Cancel the common factor.

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.2.4.1.1.4

Rewrite the expression.

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.2.4.1.2

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

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.2.4.1.3

Move the negative in front of the fraction.

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3

Solve for $C$ in $- 3 = - \frac{3}{2} + 2 C$ .

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Step 1.3.3.1

Rewrite the equation as $- \frac{3}{2} + 2 C = - 3$ .

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Add $\frac{3}{2}$ to both sides of the equation.

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2.2

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

$2 C = - 3 \cdot \frac{2}{2} + \frac{3}{2}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2.3

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

$2 C = \frac{- 3 \cdot 2}{2} + \frac{3}{2}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2.4

Combine the numerators over the common denominator.

$2 C = \frac{- 3 \cdot 2 + 3}{2}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2.5

Simplify the numerator.

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Step 1.3.3.2.5.1

Multiply $- 3$ by $2$ .

$2 C = \frac{- 6 + 3}{2}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2.5.2

Add $- 6$ and $3$ .

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.2.6

Move the negative in front of the fraction.

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.3

Divide each term in $2 C = - \frac{3}{2}$ by $2$ and simplify.

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Step 1.3.3.3.1

Divide each term in $2 C = - \frac{3}{2}$ by $2$ .

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.3.2.1.2

Divide $C$ by $1$ .

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$C = - \frac{3}{2} \cdot \frac{1}{2}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.3.3.2

Multiply $- \frac{3}{2} \cdot \frac{1}{2}$ .

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Step 1.3.3.3.3.2.1

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

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

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.3.3.3.2.2

Multiply $2$ by $2$ .

$C = - \frac{3}{4}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$C = - \frac{3}{4}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$C = - \frac{3}{4}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$C = - \frac{3}{4}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

$C = - \frac{3}{4}$

$3 = 1 + 2 B$

$A = \frac{1}{4}$

Step 1.3.4

Replace all occurrences of $C$ with $- \frac{3}{4}$ in each equation.

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Step 1.3.4.1

Rewrite the equation as $1 + 2 B = 3$ .

$1 + 2 B = 3$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.4.2

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

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Step 1.3.4.2.1

Subtract $1$ from both sides of the equation.

$2 B = 3 - 1$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.4.2.2

Subtract $1$ from $3$ .

$2 B = 2$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

$2 B = 2$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.4.3

Divide each term in $2 B = 2$ by $2$ and simplify.

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Step 1.3.4.3.1

Divide each term in $2 B = 2$ by $2$ .

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

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.4.3.2

Simplify the left side.

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Step 1.3.4.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.4.3.2.1.1

Cancel the common factor.

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

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.4.3.2.1.2

Divide $B$ by $1$ .

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

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

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

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

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

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.4.3.3

Simplify the right side.

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Step 1.3.4.3.3.1

Divide $2$ by $2$ .

$B = 1$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

$B = 1$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

$B = 1$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

$B = 1$

$C = - \frac{3}{4}$

$A = \frac{1}{4}$

Step 1.3.5

List all of the solutions.

$B = 1, C = - \frac{3}{4}, A = \frac{1}{4}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B x + C}{2 x^{2} - 3 x + 12}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{4}}{x} + \frac{1 x - \frac{3}{4}}{2 x^{2} - 3 x + 12}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator and denominator of the fraction by $4$ .

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Step 1.5.1.1

Multiply $\frac{1 \cdot x - \frac{3}{4}}{2 x^{2} - 3 x + 12}$ by $\frac{4}{4}$ .

$\frac{\frac{1}{4}}{x} + \frac{4}{4} \cdot \frac{1 \cdot x - \frac{3}{4}}{2 x^{2} - 3 x + 12}$

Step 1.5.1.2

Combine.

$\frac{\frac{1}{4}}{x} + \frac{4 (1 \cdot x - \frac{3}{4})}{4 (2 x^{2} - 3 x + 12)}$

Step 1.5.2

Apply the distributive property.

$\frac{\frac{1}{4}}{x} + \frac{4 (1 \cdot x) + 4 (- \frac{3}{4})}{4 (2 x^{2}) + 4 (- 3 x) + 4 \cdot 12}$

Step 1.5.3

Cancel the common factor of $4$ .

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Step 1.5.3.1

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$\frac{\frac{1}{4}}{x} + \frac{4 (1 \cdot x) + 4 (\frac{- 3}{4})}{4 (2 x^{2}) + 4 (- 3 x) + 4 \cdot 12}$

Step 1.5.3.2

Cancel the common factor.

$\frac{\frac{1}{4}}{x} + \frac{4 (1 \cdot x) + 4 (\frac{- 3}{4})}{4 (2 x^{2}) + 4 (- 3 x) + 4 \cdot 12}$

Step 1.5.3.3

Rewrite the expression.

$\frac{\frac{1}{4}}{x} + \frac{4 (1 \cdot x) - 3}{4 (2 x^{2}) + 4 (- 3 x) + 4 \cdot 12}$

Step 1.5.4

Multiply $4$ by $1$ .

$\frac{\frac{1}{4}}{x} + \frac{4 x - 3}{4 \cdot (2 x^{2}) + 4 (- 3 x) + 4 \cdot 12}$

Step 1.5.5

Factor $4$ out of $4 \cdot 2 x^{2} + 4 (- 3 x) + 4 \cdot 12$ .

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Step 1.5.5.1

Factor $4$ out of $4 \cdot 2 x^{2}$ .

$\frac{\frac{1}{4}}{x} + \frac{4 x - 3}{4 (2 x^{2}) + 4 (- 3 x) + 4 \cdot 12}$

Step 1.5.5.2

Factor $4$ out of $4 \cdot 12$ .

$\frac{\frac{1}{4}}{x} + \frac{4 x - 3}{4 (2 x^{2}) + 4 (- 3 x) + 4 (12)}$

Step 1.5.5.3

Factor $4$ out of $4 (2 x^{2}) + 4 (- 3 x)$ .

$\frac{\frac{1}{4}}{x} + \frac{4 x - 3}{4 (2 x^{2} - 3 x) + 4 (12)}$

Step 1.5.5.4

Factor $4$ out of $4 (2 x^{2} - 3 x) + 4 (12)$ .

$\frac{\frac{1}{4}}{x} + \frac{4 x - 3}{4 (2 x^{2} - 3 x + 12)}$

Step 1.5.6

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{x} + \frac{4 x - 3}{4 (2 x^{2} - 3 x + 12)}$

Step 1.5.7

Multiply $\frac{1}{4}$ by $\frac{1}{x}$ .

$\int \frac{1}{4 x} + \frac{4 x - 3}{4 (2 x^{2} - 3 x + 12)} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{4 x} d x + \int \frac{4 x - 3}{4 (2 x^{2} - 3 x + 12)} d x$

Step 3

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int \frac{1}{x} d x + \int \frac{4 x - 3}{4 (2 x^{2} - 3 x + 12)} d x$

Step 4

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

$\frac{1}{4} (\ln (| x |) + C) + \int \frac{4 x - 3}{4 (2 x^{2} - 3 x + 12)} d x$

Step 5

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} (\ln (| x |) + C) + \frac{1}{4} \int \frac{4 x - 3}{2 x^{2} - 3 x + 12} d x$

Step 6

Let $u = 2 x^{2} - 3 x + 12$ . Then $d u = (4 x - 3) d x$ , so $\frac{1}{4 x - 3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = 2 x^{2} - 3 x + 12$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $2 x^{2} - 3 x + 12$ .

$\frac{d}{d x} [2 x^{2} - 3 x + 12]$

Step 6.1.2

By the Sum Rule, the derivative of $2 x^{2} - 3 x + 12$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [12]$ .

$\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [12]$

Step 6.1.3

Evaluate $\frac{d}{d x} [2 x^{2}]$ .

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Step 6.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{2}$ with respect to $x$ is $2 \frac{d}{d x} [x^{2}]$ .

$2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [12]$

Step 6.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 (2 x) + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [12]$

Step 6.1.3.3

Multiply $2$ by $2$ .

$4 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [12]$

Step 6.1.4

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

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Step 6.1.4.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$4 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [12]$

Step 6.1.4.2

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

$4 x - 3 \cdot 1 + \frac{d}{d x} [12]$

Step 6.1.4.3

Multiply $- 3$ by $1$ .

$4 x - 3 + \frac{d}{d x} [12]$

Step 6.1.5

Differentiate using the Constant Rule.

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Step 6.1.5.1

Since $12$ is constant with respect to $x$ , the derivative of $12$ with respect to $x$ is $0$ .

$4 x - 3 + 0$

Step 6.1.5.2

Add $4 x - 3$ and $0$ .

$4 x - 3$

Step 6.2

Rewrite the problem using $u$ and $d u$ .

$\frac{1}{4} (\ln (| x |) + C) + \frac{1}{4} \int \frac{1}{u} d u$

Step 7

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

$\frac{1}{4} (\ln (| x |) + C) + \frac{1}{4} (\ln (| u |) + C)$

Step 8

Simplify.

$\frac{1}{4} \ln (| x |) + \frac{1}{4} \ln (| u |) + C$

Step 9

Replace all occurrences of $u$ with $2 x^{2} - 3 x + 12$ .

$\frac{1}{4} \ln (| x |) + \frac{1}{4} \ln (| 2 x^{2} - 3 x + 12 |) + C$