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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