Calculus Examples

$\int \frac{2 x^{3} + x^{2} - 21 x + 24}{x^{2} + 2 x - 8} d x$

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 $0$ .

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

Step 1.1.2

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x^{2}$ .

$2 x$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

Step 1.1.3

Multiply the new quotient term by the divisor.

$2 x$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

$2 x^{3}$

$4 x^{2}$

$16 x$

Step 1.1.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} + 4 x^{2} - 16 x$

$2 x$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

$2 x^{3}$

$4 x^{2}$

$16 x$

Step 1.1.5

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

$2 x$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

$2 x^{3}$

$4 x^{2}$

$16 x$

$3 x^{2}$

$5 x$

Step 1.1.6

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

$2 x$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

$2 x^{3}$

$4 x^{2}$

$16 x$

$3 x^{2}$

$5 x$

$24$

Step 1.1.7

Divide the highest order term in the dividend $- 3 x^{2}$ by the highest order term in divisor $x^{2}$ .

						$2 x$	-	$3$
$x^{2}$	+	$2 x$	-	$8$		$2 x^{3}$	+	$x^{2}$	-	$21 x$	+	$24$
					-	$2 x^{3}$	-	$4 x^{2}$	+	$16 x$
							-	$3 x^{2}$	-	$5 x$	+	$24$

Step 1.1.8

Multiply the new quotient term by the divisor.

						$2 x$	-	$3$
$x^{2}$	+	$2 x$	-	$8$		$2 x^{3}$	+	$x^{2}$	-	$21 x$	+	$24$
					-	$2 x^{3}$	-	$4 x^{2}$	+	$16 x$
							-	$3 x^{2}$	-	$5 x$	+	$24$
							-	$3 x^{2}$	-	$6 x$	+	$24$

Step 1.1.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 x^{2} - 6 x + 24$

$2 x$

$3$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

$2 x^{3}$

$4 x^{2}$

$16 x$

$3 x^{2}$

$5 x$

$24$

$3 x^{2}$

$6 x$

$24$

Step 1.1.10

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

$2 x$

$3$

$x^{2}$

$2 x$

$8$

$2 x^{3}$

$x^{2}$

$21 x$

$24$

$2 x^{3}$

$4 x^{2}$

$16 x$

$3 x^{2}$

$5 x$

$24$

$3 x^{2}$

$6 x$

$24$

$x$

$0$

Step 1.1.11

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

$2 x - 3 + \frac{x}{x^{2} + 2 x - 8}$

Step 1.2

Decompose the fraction and multiply through by the common denominator.

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Step 1.2.1

Factor $x^{2} + 2 x - 8$ using the AC method.

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Step 1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

Step 1.2.1.2

Write the factored form using these integers.

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

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 $A$ .

$\frac{A}{x - 2}$

Step 1.2.3

$\frac{A}{x - 2} + \frac{B}{x + 4}$

Step 1.2.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 2) (x + 4)$ .

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

Step 1.2.5

Cancel the common factor of $x - 2$ .

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Step 1.2.5.1

Cancel the common factor.

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

Step 1.2.5.2

Rewrite the expression.

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

Step 1.2.6

Cancel the common factor of $x + 4$ .

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Step 1.2.6.1

Cancel the common factor.

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

Step 1.2.6.2

Divide $x$ by $1$ .

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

Step 1.2.7

Simplify each term.

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Step 1.2.7.1

Cancel the common factor of $x - 2$ .

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Step 1.2.7.1.1

Cancel the common factor.

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

Step 1.2.7.1.2

Divide $(A) (x + 4)$ by $1$ .

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

Step 1.2.7.2

Apply the distributive property.

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

Step 1.2.7.3

Move $4$ to the left of $A$ .

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

Step 1.2.7.4

Cancel the common factor of $x + 4$ .

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Step 1.2.7.4.1

Cancel the common factor.

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

Step 1.2.7.4.2

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

$x = A x + 4 A + (B) (x - 2)$

Step 1.2.7.5

Apply the distributive property.

$x = A x + 4 A + B x + B \cdot - 2$

Step 1.2.7.6

Move $- 2$ to the left of $B$ .

$x = A x + 4 A + B x - 2 B$

Step 1.2.8

Move $4 A$ .

$x = A x + B x + 4 A - 2 B$

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 $x$ 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 the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 4 A - 2 B$

Step 1.3.3

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

$1 = A + B$

$0 = 4 A - 2 B$

$1 = A + B$

$0 = 4 A - 2 B$

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 $A + B = 1$ .

$A + B = 1$

$0 = 4 A - 2 B$

Step 1.4.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$0 = 4 A - 2 B$

$A = 1 - B$

$0 = 4 A - 2 B$

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 $0 = 4 A - 2 B$ with $1 - B$ .

$0 = 4 (1 - B) - 2 B$

$A = 1 - B$

Step 1.4.2.2

Simplify the right side.

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Step 1.4.2.2.1

Simplify $4 (1 - B) - 2 B$ .

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Step 1.4.2.2.1.1

Simplify each term.

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Step 1.4.2.2.1.1.1

Apply the distributive property.

$0 = 4 \cdot 1 + 4 (- B) - 2 B$

$A = 1 - B$

Step 1.4.2.2.1.1.2

Multiply $4$ by $1$ .

$0 = 4 + 4 (- B) - 2 B$

$A = 1 - B$

Step 1.4.2.2.1.1.3

Multiply $- 1$ by $4$ .

$0 = 4 - 4 B - 2 B$

$A = 1 - B$

$0 = 4 - 4 B - 2 B$

$A = 1 - B$

Step 1.4.2.2.1.2

Subtract $2 B$ from $- 4 B$ .

$0 = 4 - 6 B$

$A = 1 - B$

$0 = 4 - 6 B$

$A = 1 - B$

$0 = 4 - 6 B$

$A = 1 - B$

$0 = 4 - 6 B$

$A = 1 - B$

Step 1.4.3

Solve for $B$ in $0 = 4 - 6 B$ .

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Step 1.4.3.1

Rewrite the equation as $4 - 6 B = 0$ .

$4 - 6 B = 0$

$A = 1 - B$

Step 1.4.3.2

Subtract $4$ from both sides of the equation.

$- 6 B = - 4$

$A = 1 - B$

Step 1.4.3.3

Divide each term in $- 6 B = - 4$ by $- 6$ and simplify.

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Step 1.4.3.3.1

Divide each term in $- 6 B = - 4$ by $- 6$ .

$\frac{- 6 B}{- 6} = \frac{- 4}{- 6}$

$A = 1 - B$

Step 1.4.3.3.2

Simplify the left side.

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Step 1.4.3.3.2.1

Cancel the common factor of $- 6$ .

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Step 1.4.3.3.2.1.1

Cancel the common factor.

$\frac{- 6 B}{- 6} = \frac{- 4}{- 6}$

$A = 1 - B$

Step 1.4.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 4}{- 6}$

$A = 1 - B$

$B = \frac{- 4}{- 6}$

$A = 1 - B$

$B = \frac{- 4}{- 6}$

$A = 1 - B$

Step 1.4.3.3.3

Simplify the right side.

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Step 1.4.3.3.3.1

Cancel the common factor of $- 4$ and $- 6$ .

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Step 1.4.3.3.3.1.1

Factor $- 2$ out of $- 4$ .

$B = \frac{- 2 \cdot 2}{- 6}$

$A = 1 - B$

Step 1.4.3.3.3.1.2

Cancel the common factors.

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Step 1.4.3.3.3.1.2.1

Factor $- 2$ out of $- 6$ .

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

$A = 1 - B$

Step 1.4.3.3.3.1.2.2

Cancel the common factor.

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

$A = 1 - B$

Step 1.4.3.3.3.1.2.3

Rewrite the expression.

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

Step 1.4.4

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

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Step 1.4.4.1

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

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

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

Step 1.4.4.2

Simplify the right side.

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Step 1.4.4.2.1

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

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Step 1.4.4.2.1.1

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

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

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

Step 1.4.4.2.1.2

Combine the numerators over the common denominator.

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

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

Step 1.4.4.2.1.3

Subtract $2$ from $3$ .

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

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

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

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

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

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

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

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

Step 1.4.5

List all of the solutions.

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

Step 1.5

Replace each of the partial fraction coefficients in $2 x - 3 + \frac{A}{x - 2} + \frac{B}{x + 4}$ with the values found for $A$ and $B$ .

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

Step 1.6

Simplify.

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Step 1.6.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.6.2

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

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

Step 1.6.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.6.4

Multiply $\frac{2}{3}$ by $\frac{1}{x + 4}$ .

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 4

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

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

Step 5

Apply the constant rule.

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

Step 6

Combine $\frac{1}{2}$ and $x^{2}$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $x - 2$ .

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

Step 8.1.2

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

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

Step 8.1.3

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

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

Step 8.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

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

Step 9

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

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

Step 10

Since $\frac{2}{3}$ is constant with respect to $x$ , move $\frac{2}{3}$ out of the integral.

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

Step 11

Let $u_{2} = x + 4$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 11.1

Let $u_{2} = x + 4$ . Find $\frac{d u_{2}}{d x}$ .

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Step 11.1.1

Differentiate $x + 4$ .

$\frac{d}{d x} [x + 4]$

Step 11.1.2

By the Sum Rule, the derivative of $x + 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 11.1.3

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

$1 + \frac{d}{d x} [4]$

Step 11.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

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

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

Step 12

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

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

Step 13

Simplify.

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

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

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

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

Step 14.2

Replace all occurrences of $u_{2}$ with $x + 4$ .

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

Step 15

Reorder terms.

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