Calculus Examples

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

Step 1

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 1.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

Step 1.1.1

Factor the fraction.

Tap for more steps...

Step 1.1.1.1

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

Tap for more steps...

Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

Factor $x$ out of $- x$ .

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

Step 1.1.1.1.3

Factor $x$ out of $x (4 x^{2}) + x \cdot - 1$ .

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

Step 1.1.1.2

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Step 1.1.1.3

Rewrite $1$ as $1^{2}$ .

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

Step 1.1.1.4

Factor.

Tap for more steps...

Step 1.1.1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

Step 1.1.1.4.2

Remove unnecessary parentheses.

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

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 $B$ .

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.5.1

Cancel the common factor.

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $2 x + 1$ .

Tap for more steps...

Step 1.1.6.1

Cancel the common factor.

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Cancel the common factor of $2 x - 1$ .

Tap for more steps...

Step 1.1.7.1

Cancel the common factor.

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

Step 1.1.7.2

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

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

Step 1.1.8

Simplify each term.

Tap for more steps...

Step 1.1.8.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.8.1.1

Cancel the common factor.

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

Step 1.1.8.1.2

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

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

Step 1.1.8.2

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

Tap for more steps...

Step 1.1.8.2.1

Apply the distributive property.

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

Step 1.1.8.2.2

Apply the distributive property.

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

Step 1.1.8.2.3

Apply the distributive property.

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

Step 1.1.8.3

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

Tap for more steps...

Step 1.1.8.3.1

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

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

Step 1.1.8.3.2

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$4 x^{2} - 2 x + 1 = A (2 x (2 x) + 0 + 1 \cdot - 1) + \frac{(B) (x (2 x + 1) (2 x - 1))}{2 x + 1} + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.3.3

Add $2 x (2 x)$ and $0$ .

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

Step 1.1.8.4

Simplify each term.

Tap for more steps...

Step 1.1.8.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.4.2

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

Tap for more steps...

Step 1.1.8.4.2.1

Move $x$ .

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

Step 1.1.8.4.2.2

Multiply $x$ by $x$ .

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

Step 1.1.8.4.3

Multiply $2$ by $2$ .

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

Step 1.1.8.4.4

Multiply $- 1$ by $1$ .

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

Step 1.1.8.5

Apply the distributive property.

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

Step 1.1.8.6

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.7

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

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

Step 1.1.8.8

Rewrite $- 1 A$ as $- A$ .

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

Step 1.1.8.9

Cancel the common factor of $2 x + 1$ .

Tap for more steps...

Step 1.1.8.9.1

Cancel the common factor.

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

Step 1.1.8.9.2

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

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

Step 1.1.8.10

Apply the distributive property.

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

Step 1.1.8.11

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.12

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

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

Step 1.1.8.13

Simplify each term.

Tap for more steps...

Step 1.1.8.13.1

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

Tap for more steps...

Step 1.1.8.13.1.1

Move $x$ .

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

Step 1.1.8.13.1.2

Multiply $x$ by $x$ .

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

Step 1.1.8.13.2

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

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

Step 1.1.8.14

Apply the distributive property.

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

Step 1.1.8.15

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.16

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.17

Cancel the common factor of $2 x - 1$ .

Tap for more steps...

Step 1.1.8.17.1

Cancel the common factor.

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

Step 1.1.8.17.2

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

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

Step 1.1.8.18

Apply the distributive property.

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

Step 1.1.8.19

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.20

Multiply $x$ by $1$ .

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

Step 1.1.8.21

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

Tap for more steps...

Step 1.1.8.21.1

Move $x$ .

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

Step 1.1.8.21.2

Multiply $x$ by $x$ .

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

Step 1.1.8.22

Apply the distributive property.

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

Step 1.1.8.23

Rewrite using the commutative property of multiplication.

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

Step 1.1.9

Reorder.

Tap for more steps...

Step 1.1.9.1

Move $A$ .

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

Step 1.1.9.2

Move $B$ .

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

Step 1.1.9.3

Move $B$ .

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

Step 1.1.9.4

Move $- A$ .

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

Step 1.1.9.5

Move $- x B$ .

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

Step 1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

Tap for more steps...

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.

$4 = 4 A + 2 B + 2 C$

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.

$- 2 = - 1 B + C$

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.

$1 = - 1 A$

Step 1.2.4

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

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$1 = - 1 A$

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$1 = - 1 A$

Step 1.3

Solve the system of equations.

Tap for more steps...

Step 1.3.1

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

Tap for more steps...

Step 1.3.1.1

Rewrite the equation as $- 1 A = 1$ .

$- 1 A = 1$

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2

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

Tap for more steps...

Step 1.3.1.2.1

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

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

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2.2

Simplify the left side.

Tap for more steps...

Step 1.3.1.2.2.1

Dividing two negative values results in a positive value.

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

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2.2.2

Divide $A$ by $1$ .

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

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

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

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2.3

Simplify the right side.

Tap for more steps...

Step 1.3.1.2.3.1

Divide $1$ by $- 1$ .

$A = - 1$

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$A = - 1$

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$A = - 1$

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$A = - 1$

$4 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.2

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

Tap for more steps...

Step 1.3.2.1

Replace all occurrences of $A$ in $4 = 4 A + 2 B + 2 C$ with $- 1$ .

$4 = 4 (- 1) + 2 B + 2 C$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.2.2

Simplify the right side.

Tap for more steps...

Step 1.3.2.2.1

Multiply $4$ by $- 1$ .

$4 = - 4 + 2 B + 2 C$

$A = - 1$

$- 2 = - 1 B + C$

$4 = - 4 + 2 B + 2 C$

$A = - 1$

$- 2 = - 1 B + C$

$4 = - 4 + 2 B + 2 C$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3

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

Tap for more steps...

Step 1.3.3.1

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

$- 4 + 2 B + 2 C = 4$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.2

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

Tap for more steps...

Step 1.3.3.2.1

Add $4$ to both sides of the equation.

$2 B + 2 C = 4 + 4$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.2.2

Subtract $2 C$ from both sides of the equation.

$2 B = 4 + 4 - 2 C$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.2.3

Add $4$ and $4$ .

$2 B = 8 - 2 C$

$A = - 1$

$- 2 = - 1 B + C$

$2 B = 8 - 2 C$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3

Divide each term in $2 B = 8 - 2 C$ by $2$ and simplify.

Tap for more steps...

Step 1.3.3.3.1

Divide each term in $2 B = 8 - 2 C$ by $2$ .

$\frac{2 B}{2} = \frac{8}{2} + \frac{- 2 C}{2}$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.3.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{8}{2} + \frac{- 2 C}{2}$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{8}{2} + \frac{- 2 C}{2}$

$A = - 1$

$- 2 = - 1 B + C$

$B = \frac{8}{2} + \frac{- 2 C}{2}$

$A = - 1$

$- 2 = - 1 B + C$

$B = \frac{8}{2} + \frac{- 2 C}{2}$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.3.3.3.1

Simplify each term.

Tap for more steps...

Step 1.3.3.3.3.1.1

Divide $8$ by $2$ .

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

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2

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

Tap for more steps...

Step 1.3.3.3.3.1.2.1

Factor $2$ out of $- 2 C$ .

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

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2

Cancel the common factors.

Tap for more steps...

Step 1.3.3.3.3.1.2.2.1

Factor $2$ out of $2$ .

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

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2.2

Cancel the common factor.

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

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2.3

Rewrite the expression.

$B = 4 + \frac{- C}{1}$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2.4

Divide $- C$ by $1$ .

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 1 B + C$

Step 1.3.4

Replace all occurrences of $B$ with $4 - C$ in each equation.

Tap for more steps...

Step 1.3.4.1

Replace all occurrences of $B$ in $- 2 = - 1 B + C$ with $4 - C$ .

$- 2 = - 1 (4 - C) + C$

$B = 4 - C$

$A = - 1$

Step 1.3.4.2

Simplify the right side.

Tap for more steps...

Step 1.3.4.2.1

Simplify $- 1 (4 - C) + C$ .

Tap for more steps...

Step 1.3.4.2.1.1

Simplify each term.

Tap for more steps...

Step 1.3.4.2.1.1.1

Apply the distributive property.

$- 2 = - 1 \cdot 4 - 1 (- C) + C$

$B = 4 - C$

$A = - 1$

Step 1.3.4.2.1.1.2

Multiply $- 1$ by $4$ .

$- 2 = - 4 - 1 (- C) + C$

$B = 4 - C$

$A = - 1$

Step 1.3.4.2.1.1.3

Multiply $- 1 (- C)$ .

Tap for more steps...

Step 1.3.4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$- 2 = - 4 + 1 C + C$

$B = 4 - C$

$A = - 1$

Step 1.3.4.2.1.1.3.2

Multiply $C$ by $1$ .

$- 2 = - 4 + C + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 4 + C + C$

$B = 4 - C$

$A = - 1$

$- 2 = - 4 + C + C$

$B = 4 - C$

$A = - 1$

Step 1.3.4.2.1.2

Add $C$ and $C$ .

$- 2 = - 4 + 2 C$

$B = 4 - C$

$A = - 1$

$- 2 = - 4 + 2 C$

$B = 4 - C$

$A = - 1$

$- 2 = - 4 + 2 C$

$B = 4 - C$

$A = - 1$

$- 2 = - 4 + 2 C$

$B = 4 - C$

$A = - 1$

Step 1.3.5

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

Tap for more steps...

Step 1.3.5.1

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

$- 4 + 2 C = - 2$

$B = 4 - C$

$A = - 1$

Step 1.3.5.2

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

Tap for more steps...

Step 1.3.5.2.1

Add $4$ to both sides of the equation.

$2 C = - 2 + 4$

$B = 4 - C$

$A = - 1$

Step 1.3.5.2.2

Add $- 2$ and $4$ .

$2 C = 2$

$B = 4 - C$

$A = - 1$

$2 C = 2$

$B = 4 - C$

$A = - 1$

Step 1.3.5.3

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

Tap for more steps...

Step 1.3.5.3.1

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

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

$B = 4 - C$

$A = - 1$

Step 1.3.5.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.5.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.5.3.2.1.1

Cancel the common factor.

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

$B = 4 - C$

$A = - 1$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

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

$B = 4 - C$

$A = - 1$

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

$B = 4 - C$

$A = - 1$

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

$B = 4 - C$

$A = - 1$

Step 1.3.5.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.5.3.3.1

Divide $2$ by $2$ .

$C = 1$

$B = 4 - C$

$A = - 1$

$C = 1$

$B = 4 - C$

$A = - 1$

$C = 1$

$B = 4 - C$

$A = - 1$

$C = 1$

$B = 4 - C$

$A = - 1$

Step 1.3.6

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

Tap for more steps...

Step 1.3.6.1

Replace all occurrences of $C$ in $B = 4 - C$ with $1$ .

$B = 4 - (1)$

$C = 1$

$A = - 1$

Step 1.3.6.2

Simplify the right side.

Tap for more steps...

Step 1.3.6.2.1

Simplify $4 - (1)$ .

Tap for more steps...

Step 1.3.6.2.1.1

Multiply $- 1$ by $1$ .

$B = 4 - 1$

$C = 1$

$A = - 1$

Step 1.3.6.2.1.2

Subtract $1$ from $4$ .

$B = 3$

$C = 1$

$A = - 1$

$B = 3$

$C = 1$

$A = - 1$

$B = 3$

$C = 1$

$A = - 1$

$B = 3$

$C = 1$

$A = - 1$

Step 1.3.7

List all of the solutions.

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

Step 1.4

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

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

Step 1.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

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

Step 5

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

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

Step 6

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

Tap for more steps...

Step 6.1

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

Tap for more steps...

Step 6.1.1

Differentiate $2 x + 1$ .

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

Step 6.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 6.1.3

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

Tap for more steps...

Step 6.1.3.1

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

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

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 $n = 1$ .

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

Step 6.1.3.3

Multiply $2$ by $1$ .

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

Step 6.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 6.1.4.1

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

$2 + 0$

Step 6.1.4.2

Add $2$ and $0$ .

$2$

Step 6.2

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

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

Step 7

Simplify.

Tap for more steps...

Step 7.1

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

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

Step 7.2

Move $2$ to the left of $u_{1}$ .

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

Step 8

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

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

Step 9

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

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

Step 10

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 11

Let $u_{2} = 2 x - 1$ . Then $d u_{2} = 2 d x$ , so $\frac{1}{2} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 11.1

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

Tap for more steps...

Step 11.1.1

Differentiate $2 x - 1$ .

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

Step 11.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

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

Step 11.1.3

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

Tap for more steps...

Step 11.1.3.1

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

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

Step 11.1.3.2

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

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

Step 11.1.3.3

Multiply $2$ by $1$ .

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

Step 11.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 11.1.4.1

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

$2 + 0$

Step 11.1.4.2

Add $2$ and $0$ .

$2$

Step 11.2

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

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

Step 12

Simplify.

Tap for more steps...

Step 12.1

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

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

Step 12.2

Move $2$ to the left of $u_{2}$ .

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

Step 13

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

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

Step 14

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 15

Simplify.

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

Step 16

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 16.1

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

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

Step 16.2

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

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