Calculus Examples

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

Step 1

Divide

$x^{3} + x$ by

$x^{3} - 1$ .

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of

$0$ .

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$x$

$0$

Step 1.2

Divide the highest order term in the dividend

$x^{3}$ by the highest order term in divisor

$x^{3}$ .

$1$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$x$

$0$

Step 1.3

Multiply the new quotient term by the divisor.

$1$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{3}$

$0 x^{2}$

$x$

$0$

$x^{3}$

$0$

$1$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in

$x^{3} + 0 + 0 - 1$

								$1$
$x^{3}$	+	$0 x^{2}$	+	$0 x$	-	$1$		$x^{3}$	+	$0 x^{2}$	+	$x$	+	$0$
							-	$x^{3}$	-	$0$	-	$0$	+	$1$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

								$1$
$x^{3}$	+	$0 x^{2}$	+	$0 x$	-	$1$		$x^{3}$	+	$0 x^{2}$	+	$x$	+	$0$
							-	$x^{3}$	-	$0$	-	$0$	+	$1$
											+	$x$	+	$1$

Step 1.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

Apply the constant rule.

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

Step 4

Write the fraction using partial fraction decomposition.

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Step 4.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.1.1

Factor the fraction.

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Step 4.1.1.1

Rewrite

$1$ as

$1^{3}$ .

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

Step 4.1.1.2

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = x$ and

$b = 1$ .

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

Step 4.1.1.3

Simplify.

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Step 4.1.1.3.1

Multiply

$x$ by

$1$ .

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

Step 4.1.1.3.2

One to any power is one.

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

Step 4.1.2

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

$A$ .

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

Step 4.1.3

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.

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

Step 4.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is

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

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

Step 4.1.5

Cancel the common factor of

$x - 1$ .

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Step 4.1.5.1

Cancel the common factor.

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

Step 4.1.5.2

Rewrite the expression.

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

Step 4.1.6

Cancel the common factor of

$x^{2} + x + 1$ .

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Step 4.1.6.1

Cancel the common factor.

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

Step 4.1.6.2

Divide

$x + 1$ by

$1$ .

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

Step 4.1.7

Simplify each term.

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Step 4.1.7.1

Cancel the common factor of

$x - 1$ .

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Step 4.1.7.1.1

Cancel the common factor.

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

Step 4.1.7.1.2

Divide

$(A) (x^{2} + x + 1)$ by

$1$ .

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

Step 4.1.7.2

Apply the distributive property.

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

Step 4.1.7.3

Multiply

$A$ by

$1$ .

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

Step 4.1.7.4

Cancel the common factor of

$x^{2} + x + 1$ .

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Step 4.1.7.4.1

Cancel the common factor.

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

Step 4.1.7.4.2

Divide

$(B x + C) (x - 1)$ by

$1$ .

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

Step 4.1.7.5

Expand

$(B x + C) (x - 1)$ using the FOIL Method.

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Step 4.1.7.5.1

Apply the distributive property.

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

Step 4.1.7.5.2

Apply the distributive property.

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

Step 4.1.7.5.3

Apply the distributive property.

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

Step 4.1.7.6

Simplify each term.

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Step 4.1.7.6.1

Multiply

$x$ by

$x$ by adding the exponents.

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Step 4.1.7.6.1.1

Move

$x$ .

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

Step 4.1.7.6.1.2

Multiply

$x$ by

$x$ .

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

Step 4.1.7.6.2

Move

$- 1$ to the left of

$B x$ .

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

Step 4.1.7.6.3

Rewrite

$- 1 (B x)$ as

$- (B x)$ .

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

Step 4.1.7.6.4

Move

$- 1$ to the left of

$C$ .

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

Step 4.1.7.6.5

Rewrite

$- 1 C$ as

$- C$ .

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

Step 4.1.8

Simplify the expression.

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Step 4.1.8.1

Reorder

$B$ and

$x^{2}$ .

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

Step 4.1.8.2

Move

$B$ .

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

Step 4.1.8.3

Move

$A$ .

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

Step 4.1.8.4

Move

$A x$ .

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

Step 4.2

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

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Step 4.2.1

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.

$0 = A + B$

Step 4.2.2

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.

$1 = A - 1 B + C$

Step 4.2.3

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.

$1 = A - 1 C$

Step 4.2.4

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

$0 = A + B$

$1 = A - 1 B + C$

$1 = A - 1 C$

$0 = A + B$

$1 = A - 1 B + C$

$1 = A - 1 C$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Solve for

$A$ in

$0 = A + B$ .

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Step 4.3.1.1

Rewrite the equation as

$A + B = 0$ .

$A + B = 0$

$1 = A - 1 B + C$

$1 = A - 1 C$

Step 4.3.1.2

Subtract

$B$ from both sides of the equation.

$A = - B$

$1 = A - 1 B + C$

$1 = A - 1 C$

$A = - B$

$1 = A - 1 B + C$

$1 = A - 1 C$

Step 4.3.2

Replace all occurrences of

$A$ with

$- B$ in each equation.

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Step 4.3.2.1

Replace all occurrences of

$A$ in

$1 = A - 1 B + C$ with

$- B$ .

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

$A = - B$

$1 = A - 1 C$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Simplify

$(- B) - 1 B + C$ .

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Step 4.3.2.2.1.1

Rewrite

$- 1 B$ as

$- B$ .

$1 = - B - B + C$

$A = - B$

$1 = A - 1 C$

Step 4.3.2.2.1.2

Subtract

$B$ from

$- B$ .

$1 = - 2 B + C$

$A = - B$

$1 = A - 1 C$

$1 = - 2 B + C$

$A = - B$

$1 = A - 1 C$

$1 = - 2 B + C$

$A = - B$

$1 = A - 1 C$

Step 4.3.2.3

Replace all occurrences of

$A$ in

$1 = A - 1 C$ with

$- B$ .

$1 = (- B) - 1 C$

$1 = - 2 B + C$

$A = - B$

Step 4.3.2.4

Simplify the right side.

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Step 4.3.2.4.1

Rewrite

$- 1 C$ as

$- C$ .

$1 = - B - C$

$1 = - 2 B + C$

$A = - B$

$1 = - B - C$

$1 = - 2 B + C$

$A = - B$

$1 = - B - C$

$1 = - 2 B + C$

$A = - B$

Step 4.3.3

Solve for

$C$ in

$1 = - 2 B + C$ .

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Step 4.3.3.1

Rewrite the equation as

$- 2 B + C = 1$ .

$- 2 B + C = 1$

$1 = - B - C$

$A = - B$

Step 4.3.3.2

Add

$2 B$ to both sides of the equation.

$C = 1 + 2 B$

$1 = - B - C$

$A = - B$

$C = 1 + 2 B$

$1 = - B - C$

$A = - B$

Step 4.3.4

Replace all occurrences of

$C$ with

$1 + 2 B$ in each equation.

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Step 4.3.4.1

Replace all occurrences of

$C$ in

$1 = - B - C$ with

$1 + 2 B$ .

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

$C = 1 + 2 B$

$A = - B$

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

Simplify

$- B - (1 + 2 B)$ .

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Step 4.3.4.2.1.1

Simplify each term.

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Step 4.3.4.2.1.1.1

Apply the distributive property.

$1 = - B - 1 \cdot 1 - (2 B)$

$C = 1 + 2 B$

$A = - B$

Step 4.3.4.2.1.1.2

Multiply

$- 1$ by

$1$ .

$1 = - B - 1 - (2 B)$

$C = 1 + 2 B$

$A = - B$

Step 4.3.4.2.1.1.3

Multiply

$2$ by

$- 1$ .

$1 = - B - 1 - 2 B$

$C = 1 + 2 B$

$A = - B$

$1 = - B - 1 - 2 B$

$C = 1 + 2 B$

$A = - B$

Step 4.3.4.2.1.2

Subtract

$2 B$ from

$- B$ .

$1 = - 3 B - 1$

$C = 1 + 2 B$

$A = - B$

$1 = - 3 B - 1$

$C = 1 + 2 B$

$A = - B$

$1 = - 3 B - 1$

$C = 1 + 2 B$

$A = - B$

$1 = - 3 B - 1$

$C = 1 + 2 B$

$A = - B$

Step 4.3.5

Solve for

$B$ in

$1 = - 3 B - 1$ .

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Step 4.3.5.1

Rewrite the equation as

$- 3 B - 1 = 1$ .

$- 3 B - 1 = 1$

$C = 1 + 2 B$

$A = - B$

Step 4.3.5.2

Move all terms not containing

$B$ to the right side of the equation.

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Step 4.3.5.2.1

Add

$1$ to both sides of the equation.

$- 3 B = 1 + 1$

$C = 1 + 2 B$

$A = - B$

Step 4.3.5.2.2

Add

$1$ and

$1$ .

$- 3 B = 2$

$C = 1 + 2 B$

$A = - B$

$- 3 B = 2$

$C = 1 + 2 B$

$A = - B$

Step 4.3.5.3

Divide each term in

$- 3 B = 2$ by

$- 3$ and simplify.

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Step 4.3.5.3.1

Divide each term in

$- 3 B = 2$ by

$- 3$ .

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

$C = 1 + 2 B$

$A = - B$

Step 4.3.5.3.2

Simplify the left side.

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Step 4.3.5.3.2.1

Cancel the common factor of

$- 3$ .

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Step 4.3.5.3.2.1.1

Cancel the common factor.

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

$C = 1 + 2 B$

$A = - B$

Step 4.3.5.3.2.1.2

Divide

$B$ by

$1$ .

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

$C = 1 + 2 B$

$A = - B$

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

$C = 1 + 2 B$

$A = - B$

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

$C = 1 + 2 B$

$A = - B$

Step 4.3.5.3.3

Simplify the right side.

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Step 4.3.5.3.3.1

Move the negative in front of the fraction.

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

$C = 1 + 2 B$

$A = - B$

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

$C = 1 + 2 B$

$A = - B$

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

$C = 1 + 2 B$

$A = - B$

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

$C = 1 + 2 B$

$A = - B$

Step 4.3.6

Replace all occurrences of

$B$ with

$- \frac{2}{3}$ in each equation.

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Step 4.3.6.1

Replace all occurrences of

$B$ in

$C = 1 + 2 B$ with

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

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

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

$A = - B$

Step 4.3.6.2

Simplify the right side.

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Step 4.3.6.2.1

Simplify

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

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Step 4.3.6.2.1.1

Simplify each term.

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Step 4.3.6.2.1.1.1

Multiply

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

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Step 4.3.6.2.1.1.1.1

Multiply

$- 1$ by

$2$ .

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

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

$A = - B$

Step 4.3.6.2.1.1.1.2

Combine

$- 2$ and

$\frac{2}{3}$ .

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

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

$A = - B$

Step 4.3.6.2.1.1.1.3

Multiply

$- 2$ by

$2$ .

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

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

$A = - B$

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

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

$A = - B$

Step 4.3.6.2.1.1.2

Move the negative in front of the fraction.

$C = 1 - \frac{4}{3}$

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

$A = - B$

$C = 1 - \frac{4}{3}$

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

$A = - B$

Step 4.3.6.2.1.2

Simplify the expression.

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Step 4.3.6.2.1.2.1

Write

$1$ as a fraction with a common denominator.

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

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

$A = - B$

Step 4.3.6.2.1.2.2

Combine the numerators over the common denominator.

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

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

$A = - B$

Step 4.3.6.2.1.2.3

Subtract

$4$ from

$3$ .

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

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

$A = - B$

Step 4.3.6.2.1.2.4

Move the negative in front of the fraction.

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

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

$A = - B$

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

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

$A = - B$

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

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

$A = - B$

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

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

$A = - B$

Step 4.3.6.3

Replace all occurrences of

$B$ in

$A = - B$ with

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

$A = - (- \frac{2}{3})$

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

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

Step 4.3.6.4

Simplify the right side.

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Step 4.3.6.4.1

Multiply

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

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Step 4.3.6.4.1.1

Multiply

$- 1$ by

$- 1$ .

$A = 1 (\frac{2}{3})$

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

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

Step 4.3.6.4.1.2

Multiply

$\frac{2}{3}$ by

$1$ .

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

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

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

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

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

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

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

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

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

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

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

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

Step 4.3.7

List all of the solutions.

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

Step 4.4

Replace each of the partial fraction coefficients in

$\frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1}$ with the values found for

$A$ ,

$B$ , and

$C$ .

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

Step 4.5

Simplify.

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Step 4.5.1

Multiply the numerator and denominator of the fraction by

$3$ .

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Step 4.5.1.1

Multiply

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

$\frac{3}{3}$ .

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

Step 4.5.1.2

Combine.

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

Step 4.5.2

Apply the distributive property.

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

Step 4.5.3

Cancel the common factor of

$3$ .

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Step 4.5.3.1

Move the leading negative in

$- \frac{1}{3}$ into the numerator.

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

Step 4.5.3.2

Cancel the common factor.

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

Step 4.5.3.3

Rewrite the expression.

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

Step 4.5.4

Simplify the numerator.

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Step 4.5.4.1

Combine

$x$ and

$\frac{2}{3}$ .

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

Step 4.5.4.2

Cancel the common factor of

$3$ .

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Step 4.5.4.2.1

Move the leading negative in

$- \frac{x \cdot 2}{3}$ into the numerator.

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

Step 4.5.4.2.2

Cancel the common factor.

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

Step 4.5.4.2.3

Rewrite the expression.

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

Step 4.5.4.3

Multiply

$2$ by

$- 1$ .

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

Step 4.5.5

Simplify with factoring out.

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Step 4.5.5.1

Factor

$3$ out of

$3 x^{2} + 3 x + 3 \cdot 1$ .

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Step 4.5.5.1.1

Factor

$3$ out of

$3 x^{2}$ .

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

Step 4.5.5.1.2

Factor

$3$ out of

$3 x$ .

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

Step 4.5.5.1.3

Factor

$3$ out of

$3 \cdot 1$ .

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

Step 4.5.5.1.4

Factor

$3$ out of

$3 (x^{2}) + 3 (x)$ .

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

Step 4.5.5.1.5

Factor

$3$ out of

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

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

Step 4.5.5.2

Factor

$- 1$ out of

$- 2 x$ .

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

Step 4.5.5.3

Rewrite

$- 1$ as

$- 1 (1)$ .

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

Step 4.5.5.4

Factor

$- 1$ out of

$- (2 x) - 1 (1)$ .

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

Step 4.5.5.5

Simplify the expression.

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Step 4.5.5.5.1

Rewrite

$- (2 x + 1)$ as

$- 1 (2 x + 1)$ .

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

Step 4.5.5.5.2

Move the negative in front of the fraction.

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

Step 4.5.6

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.7

Multiply

$\frac{2}{3}$ by

$\frac{1}{x - 1}$ .

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

Since

$\frac{2}{3}$ is constant with respect to

$x$ , move

$\frac{2}{3}$ out of the integral.

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

Step 7

Let

$u_{1} = x - 1$ . Then

$d u_{1} = d x$ . Rewrite using

$u_{1}$ and

$d$

$u_{1}$ .

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Step 7.1

Let

$u_{1} = x - 1$ . Find

$\frac{d u_{1}}{d x}$ .

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Step 7.1.1

Differentiate

$x - 1$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- 1$ with respect to

$x$ is

$0$ .

$1 + 0$

Step 7.1.5

Add

$1$ and

$0$ .

$1$

Step 7.2

Rewrite the problem using

$u_{1}$ and

$d u_{1}$ .

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

Step 8

The integral of

$\frac{1}{u_{1}}$ with respect to

$u_{1}$ is

$\ln (| u_{1} |)$ .

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

Step 9

Since

$- 1$ is constant with respect to

$x$ , move

$- 1$ out of the integral.

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

Step 10

Since

$\frac{1}{3}$ is constant with respect to

$x$ , move

$\frac{1}{3}$ out of the integral.

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

Step 11

Let

$u_{2} = x^{2} + x + 1$ . Then

$d u_{2} = (2 x + 1) d x$ , so

$\frac{1}{2 x + 1} d u_{2} = d x$ . Rewrite using

$u_{2}$ and

$d$

$u_{2}$ .

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Step 11.1

Let

$u_{2} = x^{2} + x + 1$ . Find

$\frac{d u_{2}}{d x}$ .

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Step 11.1.1

Differentiate

$x^{2} + x + 1$ .

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

Step 11.1.2

By the Sum Rule, the derivative of

$x^{2} + x + 1$ with respect to

$x$ is

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

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

Step 11.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 2$ .

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

Step 11.1.4

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 11.1.5

Since

$1$ is constant with respect to

$x$ , the derivative of

$1$ with respect to

$x$ is

$0$ .

$2 x + 1 + 0$

Step 11.1.6

Add

$2 x + 1$ and

$0$ .

$2 x + 1$

Step 11.2

Rewrite the problem using

$u_{2}$ and

$d u_{2}$ .

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

Step 12

The integral of

$\frac{1}{u_{2}}$ with respect to

$u_{2}$ is

$\ln (| u_{2} |)$ .

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

Step 13

Simplify.

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

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

Replace all occurrences of

$u_{1}$ with

$x - 1$ .

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

Step 14.2

Replace all occurrences of

$u_{2}$ with

$x^{2} + x + 1$ .

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

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