Calculus Examples

$\int \frac{x^{3} - 6 x - 4}{x + 2} d x$

Step 1

Let $u = x + 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

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

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Step 1.1.1

Differentiate $x + 2$ .

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

Step 1.1.2

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 1.1.3

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 1.1.4

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

$1 + 0$

Step 1.1.5

Add $1$ and $0$ .

$1$

Step 1.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{{(u - 2)}^{3} - 6 (u - 2) - 4}{u} d u$

Step 2

Split the fraction into multiple fractions.

$\int \frac{{(u - 2)}^{3}}{u} + \frac{- 6 (u - 2)}{u} + \frac{- 4}{u} d u$

Step 3

Split the single integral into multiple integrals.

$\int \frac{{(u - 2)}^{3}}{u} d u + \int \frac{- 6 (u - 2)}{u} d u + \int \frac{- 4}{u} d u$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$\int \frac{{(u - 2)}^{3}}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int \frac{- 4}{u} d u$

Step 4.2

Move the negative in front of the fraction.

$\int \frac{{(u - 2)}^{3}}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 5

Use the Binomial Theorem.

$\int \frac{u^{3} + 3 u^{2} \cdot - 2 + 3 u {(- 2)}^{2} + {(- 2)}^{3}}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6

Simplify the expression.

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Step 6.1

Rewrite the exponentiation as a product.

$\int \frac{u^{3} + 3 u^{2} \cdot - 2 + 3 u (- 2 \cdot - 2) + {(- 2)}^{3}}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.2

Rewrite the exponentiation as a product.

$\int \frac{u^{3} + 3 u^{2} \cdot - 2 + 3 u (- 2 \cdot - 2) - 2 {(- 2)}^{2}}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.3

Rewrite the exponentiation as a product.

$\int \frac{u^{3} + 3 u^{2} \cdot - 2 + 3 u (- 2 \cdot - 2) - 2 (- 2 \cdot - 2)}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.4

Move $u^{2}$ .

$\int \frac{u^{3} + 3 \cdot - 2 u^{2} + 3 u (- 2 \cdot - 2) - 2 (- 2 \cdot - 2)}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.5

Move $u$ .

$\int \frac{u^{3} + 3 \cdot - 2 u^{2} + 3 \cdot - 2 \cdot - 2 u - 2 (- 2 \cdot - 2)}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.6

Multiply $3$ by $- 2$ .

$\int \frac{u^{3} - 6 u^{2} + 3 \cdot - 2 \cdot - 2 u - 2 \cdot - 2 \cdot - 2}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.7

Multiply $3$ by $- 2$ .

$\int \frac{u^{3} - 6 u^{2} - 6 \cdot - 2 u - 2 \cdot - 2 \cdot - 2}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.8

Multiply $- 6$ by $- 2$ .

$\int \frac{u^{3} - 6 u^{2} + 12 u - 2 \cdot - 2 \cdot - 2}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.9

Multiply $- 2$ by $- 2$ .

$\int \frac{u^{3} - 6 u^{2} + 12 u + 4 \cdot - 2}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 6.10

Multiply $4$ by $- 2$ .

$\int \frac{u^{3} - 6 u^{2} + 12 u - 8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 7

Divide $u^{3} - 6 u^{2} + 12 u - 8$ by $u$ .

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Step 7.1

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

$u$

$0$

$u^{3}$

$6 u^{2}$

$12 u$

$8$

Step 7.2

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

					$u^{2}$
	$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$

Step 7.3

Multiply the new quotient term by the divisor.

				$u^{2}$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			+	$u^{3}$	+	$0$

Step 7.4

The expression needs to be subtracted from the dividend, so change all the signs in $u^{3} + 0$

				$u^{2}$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$

Step 7.5

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

				$u^{2}$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$

Step 7.6

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

				$u^{2}$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$

Step 7.7

Divide the highest order term in the dividend $- 6 u^{2}$ by the highest order term in divisor $u$ .

				$u^{2}$	-	$6 u$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$

Step 7.8

Multiply the new quotient term by the divisor.

				$u^{2}$	-	$6 u$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					-	$6 u^{2}$	+	$0$

Step 7.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 u^{2} + 0$

				$u^{2}$	-	$6 u$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$

Step 7.10

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

				$u^{2}$	-	$6 u$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$
							+	$12 u$

Step 7.11

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

				$u^{2}$	-	$6 u$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$
							+	$12 u$	-	$8$

Step 7.12

Divide the highest order term in the dividend $12 u$ by the highest order term in divisor $u$ .

				$u^{2}$	-	$6 u$	+	$12$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$
							+	$12 u$	-	$8$

Step 7.13

Multiply the new quotient term by the divisor.

				$u^{2}$	-	$6 u$	+	$12$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$
							+	$12 u$	-	$8$
							+	$12 u$	+	$0$

Step 7.14

The expression needs to be subtracted from the dividend, so change all the signs in $12 u + 0$

				$u^{2}$	-	$6 u$	+	$12$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$
							+	$12 u$	-	$8$
							-	$12 u$	-	$0$

Step 7.15

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

				$u^{2}$	-	$6 u$	+	$12$
$u$	+	$0$		$u^{3}$	-	$6 u^{2}$	+	$12 u$	-	$8$
			-	$u^{3}$	-	$0$
					-	$6 u^{2}$	+	$12 u$
					+	$6 u^{2}$	-	$0$
							+	$12 u$	-	$8$
							-	$12 u$	-	$0$
									-	$8$

Step 7.16

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

$\int u^{2} - 6 u + 12 - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 8

Split the single integral into multiple integrals.

$\int u^{2} d u + \int - 6 u d u + \int 12 d u + \int - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 9

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

$\frac{1}{3} u^{3} + C + \int - 6 u d u + \int 12 d u + \int - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 10

Since $- 6$ is constant with respect to $u$ , move $- 6$ out of the integral.

$\frac{1}{3} u^{3} + C - 6 \int u d u + \int 12 d u + \int - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 11

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

$\frac{1}{3} u^{3} + C - 6 (\frac{1}{2} u^{2} + C) + \int 12 d u + \int - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 12

Apply the constant rule.

$\frac{1}{3} u^{3} + C - 6 (\frac{1}{2} u^{2} + C) + 12 u + C + \int - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 13

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C + \int - \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 14

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - \int \frac{8}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 15

Since $8$ is constant with respect to $u$ , move $8$ out of the integral.

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - (8 \int \frac{1}{u} d u) + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 16

Multiply $8$ by $- 1$ .

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 \int \frac{1}{u} d u + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 17

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) + \int - \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 18

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - \int \frac{6 (u - 2)}{u} d u + \int - \frac{4}{u} d u$

Step 19

Since $6$ is constant with respect to $u$ , move $6$ out of the integral.

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - (6 \int \frac{u - 2}{u} d u) + \int - \frac{4}{u} d u$

Step 20

Multiply $6$ by $- 1$ .

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 \int \frac{u - 2}{u} d u + \int - \frac{4}{u} d u$

Step 21

Divide $u - 2$ by $u$ .

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Step 21.1

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

$u$

$0$

$u$

$2$

Step 21.2

Divide the highest order term in the dividend $u$ by the highest order term in divisor $u$ .

					$1$
	$u$	+	$0$		$u$	-	$2$

Step 21.3

Multiply the new quotient term by the divisor.

				$1$
$u$	+	$0$		$u$	-	$2$
			+	$u$	+	$0$

Step 21.4

The expression needs to be subtracted from the dividend, so change all the signs in $u + 0$

				$1$
$u$	+	$0$		$u$	-	$2$
			-	$u$	-	$0$

Step 21.5

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

				$1$
$u$	+	$0$		$u$	-	$2$
			-	$u$	-	$0$
					-	$2$

Step 21.6

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 \int 1 - \frac{2}{u} d u + \int - \frac{4}{u} d u$

Step 22

Split the single integral into multiple integrals.

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (\int d u + \int - \frac{2}{u} d u) + \int - \frac{4}{u} d u$

Step 23

Apply the constant rule.

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C + \int - \frac{2}{u} d u) + \int - \frac{4}{u} d u$

Step 24

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - \int \frac{2}{u} d u) + \int - \frac{4}{u} d u$

Step 25

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - (2 \int \frac{1}{u} d u)) + \int - \frac{4}{u} d u$

Step 26

Multiply $2$ by $- 1$ .

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - 2 \int \frac{1}{u} d u) + \int - \frac{4}{u} d u$

Step 27

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - 2 (\ln (| u |) + C)) + \int - \frac{4}{u} d u$

Step 28

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - 2 (\ln (| u |) + C)) - \int \frac{4}{u} d u$

Step 29

Since $4$ is constant with respect to $u$ , move $4$ out of the integral.

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - 2 (\ln (| u |) + C)) - (4 \int \frac{1}{u} d u)$

Step 30

Multiply $4$ by $- 1$ .

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - 2 (\ln (| u |) + C)) - 4 \int \frac{1}{u} d u$

Step 31

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

$\frac{1}{3} u^{3} + C - 6 (\frac{u^{2}}{2} + C) + 12 u + C - 8 (\ln (| u |) + C) - 6 (u + C - 2 (\ln (| u |) + C)) - 4 (\ln (| u |) + C)$

Step 32

Simplify.

$\frac{u^{3}}{3} - 3 u^{2} + 12 u - 8 \ln (| u |) - 6 u + 12 \ln (| u |) - 4 \ln (| u |) + C$

Step 33

Reorder terms.

$\frac{1}{3} u^{3} - 3 u^{2} + 12 u - 8 \ln (| u |) - 6 u + 12 \ln (| u |) - 4 \ln (| u |) + C$

Step 34

Simplify.

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Step 34.1

Subtract $6 u$ from $12 u$ .

$\frac{1}{3} u^{3} - 3 u^{2} + 6 u - 8 \ln (| u |) + 12 \ln (| u |) - 4 \ln (| u |) + C$

Step 34.2

Add $- 8 \ln (| u |)$ and $12 \ln (| u |)$ .

$\frac{1}{3} u^{3} - 3 u^{2} + 6 u + 4 \ln (| u |) - 4 \ln (| u |) + C$

Step 34.3

Subtract $4 \ln (| u |)$ from $4 \ln (| u |)$ .

$\frac{1}{3} u^{3} - 3 u^{2} + 6 u + 0 + C$

Step 34.4

Add $\frac{1}{3} u^{3} - 3 u^{2} + 6 u$ and $0$ .

$\frac{1}{3} u^{3} - 3 u^{2} + 6 u + C$

Step 35

Replace all occurrences of $u$ with $x + 2$ .

$\frac{1}{3} {(x + 2)}^{3} - 3 {(x + 2)}^{2} + 6 (x + 2) + C$