Calculus Examples

$\int \frac{x^{5}}{x^{3} - 1} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Divide using long polynomial division.

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Step 1.1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

Step 1.1.2

Divide the highest order term in the dividend $x^{5}$ by the highest order term in divisor $x^{3}$ .

$x^{2}$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

Step 1.1.3

Multiply the new quotient term by the divisor.

$x^{2}$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{2}$

Step 1.1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{5} + 0 + 0 - x^{2}$

$x^{2}$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{2}$

Step 1.1.5

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

$x^{2}$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{2}$

Step 1.1.6

Pull the next terms from the original dividend down into the current dividend.

$x^{2}$

$x^{3}$

$0 x^{2}$

$0 x$

$1$

$x^{5}$

$0 x^{4}$

$0 x^{3}$

$0 x^{2}$

$0 x$

$0$

$x^{5}$

$0$

$x^{2}$

$0 x$

$0$

Step 1.1.7

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

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

Step 1.2

Decompose the fraction and multiply through by the common denominator.

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Step 1.2.1

Factor the fraction.

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Step 1.2.1.1

Rewrite $1$ as $1^{3}$ .

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

Step 1.2.1.2

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 $b = 1$ .

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

Step 1.2.1.3

Simplify.

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Step 1.2.1.3.1

Multiply $x$ by $1$ .

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

Step 1.2.1.3.2

One to any power is one.

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

Step 1.2.2

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 1.2.3

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 1.2.4

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

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

Step 1.2.5

Cancel the common factor of $x - 1$ .

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Step 1.2.5.1

Cancel the common factor.

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

Step 1.2.5.2

Rewrite the expression.

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

Step 1.2.6

Cancel the common factor of $x^{2} + x + 1$ .

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Step 1.2.6.1

Cancel the common factor.

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

Step 1.2.6.2

Divide $x^{2}$ by $1$ .

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

Step 1.2.7

Simplify each term.

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Step 1.2.7.1

Cancel the common factor of $x - 1$ .

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Step 1.2.7.1.1

Cancel the common factor.

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

Step 1.2.7.1.2

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

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

Step 1.2.7.2

Apply the distributive property.

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

Step 1.2.7.3

Multiply $A$ by $1$ .

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

Step 1.2.7.4

Cancel the common factor of $x^{2} + x + 1$ .

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Step 1.2.7.4.1

Cancel the common factor.

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

Step 1.2.7.4.2

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

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

Step 1.2.7.5

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

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Step 1.2.7.5.1

Apply the distributive property.

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

Step 1.2.7.5.2

Apply the distributive property.

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

Step 1.2.7.5.3

Apply the distributive property.

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

Step 1.2.7.6

Simplify each term.

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Step 1.2.7.6.1

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

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Step 1.2.7.6.1.1

Move $x$ .

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

Step 1.2.7.6.1.2

Multiply $x$ by $x$ .

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

Step 1.2.7.6.2

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

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

Step 1.2.7.6.3

Rewrite $- 1 (B x)$ as $- (B x)$ .

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

Step 1.2.7.6.4

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

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

Step 1.2.7.6.5

Rewrite $- 1 C$ as $- C$ .

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

Step 1.2.8

Simplify the expression.

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Step 1.2.8.1

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

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

Step 1.2.8.2

Move $B$ .

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

Step 1.2.8.3

Move $A$ .

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

Step 1.2.8.4

Move $A x$ .

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

Step 1.3

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

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Step 1.3.1

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.

$1 = A + B$

Step 1.3.2

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.

$0 = A - 1 B + C$

Step 1.3.3

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.

$0 = A - 1 C$

Step 1.3.4

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

$1 = A + B$

$0 = A - 1 B + C$

$0 = A - 1 C$

$1 = A + B$

$0 = A - 1 B + C$

$0 = A - 1 C$

Step 1.4

Solve the system of equations.

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Step 1.4.1

Solve for $A$ in $1 = A + B$ .

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Step 1.4.1.1

Rewrite the equation as $A + B = 1$ .

$A + B = 1$

$0 = A - 1 B + C$

$0 = A - 1 C$

Step 1.4.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$0 = A - 1 B + C$

$0 = A - 1 C$

$A = 1 - B$

$0 = A - 1 B + C$

$0 = A - 1 C$

Step 1.4.2

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

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Step 1.4.2.1

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

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

$A = 1 - B$

$0 = A - 1 C$

Step 1.4.2.2

Simplify the right side.

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Step 1.4.2.2.1

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

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Step 1.4.2.2.1.1

Rewrite $- 1 B$ as $- B$ .

$0 = 1 - B - B + C$

$A = 1 - B$

$0 = A - 1 C$

Step 1.4.2.2.1.2

Subtract $B$ from $- B$ .

$0 = 1 - 2 B + C$

$A = 1 - B$

$0 = A - 1 C$

$0 = 1 - 2 B + C$

$A = 1 - B$

$0 = A - 1 C$

$0 = 1 - 2 B + C$

$A = 1 - B$

$0 = A - 1 C$

Step 1.4.2.3

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

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

$0 = 1 - 2 B + C$

$A = 1 - B$

Step 1.4.2.4

Simplify the right side.

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Step 1.4.2.4.1

Rewrite $- 1 C$ as $- C$ .

$0 = 1 - B - C$

$0 = 1 - 2 B + C$

$A = 1 - B$

$0 = 1 - B - C$

$0 = 1 - 2 B + C$

$A = 1 - B$

$0 = 1 - B - C$

$0 = 1 - 2 B + C$

$A = 1 - B$

Step 1.4.3

Reorder $1$ and $- B$ .

$A = - B + 1$

$0 = 1 - B - C$

$0 = 1 - 2 B + C$

Step 1.4.4

Solve for $C$ in $0 = 1 - 2 B + C$ .

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Step 1.4.4.1

Rewrite the equation as $1 - 2 B + C = 0$ .

$1 - 2 B + C = 0$

$A = - B + 1$

$0 = 1 - B - C$

Step 1.4.4.2

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

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Step 1.4.4.2.1

Subtract $1$ from both sides of the equation.

$- 2 B + C = - 1$

$A = - B + 1$

$0 = 1 - B - C$

Step 1.4.4.2.2

Add $2 B$ to both sides of the equation.

$C = - 1 + 2 B$

$A = - B + 1$

$0 = 1 - B - C$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = 1 - B - C$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = 1 - B - C$

Step 1.4.5

Replace all occurrences of $C$ with $- 1 + 2 B$ in each equation.

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Step 1.4.5.1

Replace all occurrences of $C$ in $0 = 1 - B - C$ with $- 1 + 2 B$ .

$0 = 1 - B - (- 1 + 2 B)$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.5.2

Simplify the right side.

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Step 1.4.5.2.1

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

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Step 1.4.5.2.1.1

Simplify each term.

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Step 1.4.5.2.1.1.1

Apply the distributive property.

$0 = 1 - B + 1 - (2 B)$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.5.2.1.1.2

Multiply $- 1$ by $- 1$ .

$0 = 1 - B + 1 - (2 B)$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.5.2.1.1.3

Multiply $2$ by $- 1$ .

$0 = 1 - B + 1 - 2 B$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = 1 - B + 1 - 2 B$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.5.2.1.2

Simplify by adding terms.

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Step 1.4.5.2.1.2.1

Add $1$ and $1$ .

$0 = - B + 2 - 2 B$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.5.2.1.2.2

Subtract $2 B$ from $- B$ .

$0 = - 3 B + 2$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = - 3 B + 2$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = - 3 B + 2$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = - 3 B + 2$

$C = - 1 + 2 B$

$A = - B + 1$

$0 = - 3 B + 2$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.6

Solve for $B$ in $0 = - 3 B + 2$ .

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Step 1.4.6.1

Rewrite the equation as $- 3 B + 2 = 0$ .

$- 3 B + 2 = 0$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.6.2

Subtract $2$ from both sides of the equation.

$- 3 B = - 2$

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.6.3

Divide each term in $- 3 B = - 2$ by $- 3$ and simplify.

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Step 1.4.6.3.1

Divide each term in $- 3 B = - 2$ by $- 3$ .

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

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.6.3.2

Simplify the left side.

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Step 1.4.6.3.2.1

Cancel the common factor of $- 3$ .

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Step 1.4.6.3.2.1.1

Cancel the common factor.

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

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.6.3.2.1.2

Divide $B$ by $1$ .

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

$C = - 1 + 2 B$

$A = - B + 1$

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

$C = - 1 + 2 B$

$A = - B + 1$

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

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.6.3.3

Simplify the right side.

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Step 1.4.6.3.3.1

Dividing two negative values results in a positive value.

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

$C = - 1 + 2 B$

$A = - B + 1$

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

$C = - 1 + 2 B$

$A = - B + 1$

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

$C = - 1 + 2 B$

$A = - B + 1$

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

$C = - 1 + 2 B$

$A = - B + 1$

Step 1.4.7

Replace all occurrences of $B$ with $\frac{2}{3}$ in each equation.

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Step 1.4.7.1

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

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

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

$A = - B + 1$

Step 1.4.7.2

Simplify the right side.

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Step 1.4.7.2.1

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

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Step 1.4.7.2.1.1

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

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Step 1.4.7.2.1.1.1

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

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

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

$A = - B + 1$

Step 1.4.7.2.1.1.2

Multiply $2$ by $2$ .

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

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

$A = - B + 1$

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

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

$A = - B + 1$

Step 1.4.7.2.1.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

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

$A = - B + 1$

Step 1.4.7.2.1.3

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

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

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

$A = - B + 1$

Step 1.4.7.2.1.4

Combine the numerators over the common denominator.

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

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

$A = - B + 1$

Step 1.4.7.2.1.5

Simplify the numerator.

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Step 1.4.7.2.1.5.1

Multiply $- 1$ by $3$ .

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

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

$A = - B + 1$

Step 1.4.7.2.1.5.2

Add $- 3$ and $4$ .

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

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

$A = - B + 1$

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

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

$A = - B + 1$

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

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

$A = - B + 1$

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

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

$A = - B + 1$

Step 1.4.7.3

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

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

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

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

Step 1.4.7.4

Simplify the right side.

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Step 1.4.7.4.1

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

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Step 1.4.7.4.1.1

Write $1$ as a fraction with a common denominator.

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

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

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

Step 1.4.7.4.1.2

Combine the numerators over the common denominator.

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

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

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

Step 1.4.7.4.1.3

Add $- 2$ and $3$ .

$A = \frac{1}{3}$

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

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

$A = \frac{1}{3}$

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

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

$A = \frac{1}{3}$

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

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

$A = \frac{1}{3}$

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

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

Step 1.4.8

List all of the solutions.

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

Step 1.5

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

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

Step 1.6

Simplify.

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Step 1.6.1

Multiply the numerator and denominator of the fraction by $3$ .

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Step 1.6.1.1

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

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

Step 1.6.1.2

Combine.

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

Step 1.6.2

Apply the distributive property.

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

Step 1.6.3

Simplify terms.

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Step 1.6.3.1

Cancel the common factor of $3$ .

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Step 1.6.3.1.1

Cancel the common factor.

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

Step 1.6.3.1.2

Rewrite the expression.

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

Step 1.6.3.2

Cancel the common factor of $3$ .

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Step 1.6.3.2.1

Cancel the common factor.

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

Step 1.6.3.2.2

Rewrite the expression.

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

Step 1.6.3.3

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

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Step 1.6.3.3.1

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

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

Step 1.6.3.3.2

Factor $3$ out of $3 x$ .

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

Step 1.6.3.3.3

Factor $3$ out of $3 \cdot 1$ .

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

Step 1.6.3.3.4

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

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

Step 1.6.3.3.5

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

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

Step 1.6.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.6.5

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

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

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

Step 4

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

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

Step 5

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

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Step 5.1

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

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Step 5.1.1

Differentiate $x - 1$ .

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

Step 5.1.2

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 5.1.3

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 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

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

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

Step 6

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

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

Step 7

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

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

Step 8

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 8.1

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

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Step 8.1.1

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

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

Step 8.1.2

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 8.1.3

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 8.1.4

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 8.1.5

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

$2 x + 1 + 0$

Step 8.1.6

Add $2 x + 1$ and $0$ .

$2 x + 1$

Step 8.2

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

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

Step 9

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

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

Step 10

Simplify.

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

Step 11

Substitute back in for each integration substitution variable.

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Step 11.1

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

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

Step 11.2

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

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