Calculus Examples

$\int \frac{12 x^{2}}{2 x + 1} d x$

Step 1

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$12 \int \frac{x^{2}}{2 x + 1} d x$

Step 2

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

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Step 2.1

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

$2 x$

$1$

$x^{2}$

$0 x$

$0$

Step 2.2

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

					$\frac{x}{2}$
	$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$

Step 2.3

Multiply the new quotient term by the divisor.

				$\frac{x}{2}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			+	$x^{2}$	+	$\frac{x}{2}$

Step 2.4

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

				$\frac{x}{2}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$

Step 2.5

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

				$\frac{x}{2}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$
					-	$\frac{x}{2}$

Step 2.6

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

				$\frac{x}{2}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$
					-	$\frac{x}{2}$	+	$0$

Step 2.7

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

				$\frac{x}{2}$	-	$\frac{1}{4}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$
					-	$\frac{x}{2}$	+	$0$

Step 2.8

Multiply the new quotient term by the divisor.

				$\frac{x}{2}$	-	$\frac{1}{4}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$
					-	$\frac{x}{2}$	+	$0$
					-	$\frac{x}{2}$	-	$\frac{1}{4}$

Step 2.9

The expression needs to be subtracted from the dividend, so change all the signs in $- \frac{x}{2} - \frac{1}{4}$

				$\frac{x}{2}$	-	$\frac{1}{4}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$
					-	$\frac{x}{2}$	+	$0$
					+	$\frac{x}{2}$	+	$\frac{1}{4}$

Step 2.10

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

				$\frac{x}{2}$	-	$\frac{1}{4}$
$2 x$	+	$1$		$x^{2}$	+	$0 x$	+	$0$
			-	$x^{2}$	-	$\frac{x}{2}$
					-	$\frac{x}{2}$	+	$0$
					+	$\frac{x}{2}$	+	$\frac{1}{4}$
							+	$\frac{1}{4}$

Step 2.11

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

$12 \int \frac{x}{2} - \frac{1}{4} + \frac{1}{4 (2 x + 1)} d x$

Step 3

Split the single integral into multiple integrals.

$12 (\int \frac{x}{2} d x + \int - \frac{1}{4} d x + \int \frac{1}{4 (2 x + 1)} d x)$

Step 4

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$12 (\frac{1}{2} \int x d x + \int - \frac{1}{4} d x + \int \frac{1}{4 (2 x + 1)} d x)$

Step 5

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

$12 (\frac{1}{2} (\frac{1}{2} x^{2} + C) + \int - \frac{1}{4} d x + \int \frac{1}{4 (2 x + 1)} d x)$

Step 6

Apply the constant rule.

$12 (\frac{1}{2} (\frac{1}{2} x^{2} + C) - \frac{1}{4} x + C + \int \frac{1}{4 (2 x + 1)} d x)$

Step 7

Simplify.

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Step 7.1

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

$12 (\frac{1}{2} (\frac{x^{2}}{2} + C) - \frac{1}{4} x + C + \int \frac{1}{4 (2 x + 1)} d x)$

Step 7.2

Combine $x$ and $\frac{1}{4}$ .

$12 (\frac{1}{2} (\frac{x^{2}}{2} + C) - \frac{x}{4} + C + \int \frac{1}{4 (2 x + 1)} d x)$

Step 8

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$12 (\frac{1}{2} (\frac{x^{2}}{2} + C) - \frac{x}{4} + C + \frac{1}{4} \int \frac{1}{2 x + 1} d x)$

Step 9

Let $u = 2 x + 1$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 9.1

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

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Step 9.1.1

Differentiate $2 x + 1$ .

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

Step 9.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 9.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 9.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 9.1.3.2

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

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

Step 9.1.3.3

Multiply $2$ by $1$ .

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

Step 9.1.4

Differentiate using the Constant Rule.

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Step 9.1.4.1

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

$2 + 0$

Step 9.1.4.2

Add $2$ and $0$ .

$2$

Step 9.2

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

$12 (\frac{1}{2} (\frac{x^{2}}{2} + C) - \frac{x}{4} + C + \frac{1}{4} \int \frac{1}{u} \cdot \frac{1}{2} d u)$

Step 10

Simplify.

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Step 10.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$12 (\frac{1}{2} (\frac{x^{2}}{2} + C) - \frac{x}{4} + C + \frac{1}{4} \int \frac{1}{u \cdot 2} d u)$

Step 10.2

Move $2$ to the left of $u$ .

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

Step 11

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

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

Step 12

Simplify.

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Step 12.1

Multiply $\frac{1}{2}$ by $\frac{1}{4}$ .

$12 (\frac{1}{2} (\frac{x^{2}}{2} + C) - \frac{x}{4} + C + \frac{1}{2 \cdot 4} \int \frac{1}{u} d u)$

Step 12.2

Multiply $2$ by $4$ .

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

Step 13

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

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

Step 14

Simplify.

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

Step 15

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 16

Simplify.

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Step 16.1

To write $\frac{x^{2}}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$12 (- \frac{x}{4} + \frac{x^{2}}{4} \cdot \frac{2}{2} + \frac{\ln (| 2 x + 1 |)}{8}) + C$

Step 16.2

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

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Step 16.2.1

Multiply $\frac{x^{2}}{4}$ by $\frac{2}{2}$ .

$12 (- \frac{x}{4} + \frac{x^{2} \cdot 2}{4 \cdot 2} + \frac{\ln (| 2 x + 1 |)}{8}) + C$

Step 16.2.2

Multiply $4$ by $2$ .

$12 (- \frac{x}{4} + \frac{x^{2} \cdot 2}{8} + \frac{\ln (| 2 x + 1 |)}{8}) + C$

Step 16.3

Combine the numerators over the common denominator.

$12 (- \frac{x}{4} + \frac{x^{2} \cdot 2 + \ln (| 2 x + 1 |)}{8}) + C$

Step 16.4

Move $2$ to the left of $x^{2}$ .

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

Step 16.5

Apply the distributive property.

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

Step 16.6

Cancel the common factor of $4$ .

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Step 16.6.1

Move the leading negative in $- \frac{x}{4}$ into the numerator.

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

Step 16.6.2

Factor $4$ out of $12$ .

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

Step 16.6.3

Cancel the common factor.

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

Step 16.6.4

Rewrite the expression.

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

Step 16.7

Multiply $- 1$ by $3$ .

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

Step 16.8

Cancel the common factor of $4$ .

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Step 16.8.1

Factor $4$ out of $12$ .

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

Step 16.8.2

Factor $4$ out of $8$ .

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

Step 16.8.3

Cancel the common factor.

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

Step 16.8.4

Rewrite the expression.

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

Step 16.9

Combine $3$ and $\frac{2 x^{2} + \ln (| 2 x + 1 |)}{2}$ .

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

Step 16.10

To write $- 3 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 16.11

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

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

Step 16.12

Combine the numerators over the common denominator.

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

Step 16.13

Simplify the numerator.

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Step 16.13.1

Factor $3$ out of $- 3 x \cdot 2 + 3 (2 x^{2} + \ln (| 2 x + 1 |))$ .

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Step 16.13.1.1

Factor $3$ out of $- 3 x \cdot 2$ .

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

Step 16.13.1.2

Factor $3$ out of $3 (- x \cdot 2) + 3 (2 x^{2} + \ln (| 2 x + 1 |))$ .

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

Step 16.13.2

Multiply $2$ by $- 1$ .

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

Step 16.14

Factor $- 1$ out of $- 2 x$ .

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

Step 16.15

Factor $- 1$ out of $2 x^{2}$ .

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

Step 16.16

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

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

Step 16.17

Factor $- 1$ out of $\ln (| 2 x + 1 |)$ .

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

Step 16.18

Factor $- 1$ out of $- (2 x - 2 x^{2}) - 1 (- \ln (| 2 x + 1 |))$ .

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

Step 16.19

Rewrite $- (2 x - 2 x^{2} - \ln (| 2 x + 1 |))$ as $- 1 (2 x - 2 x^{2} - \ln (| 2 x + 1 |))$ .

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

Step 16.20

Move the negative in front of the fraction.

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

Step 17

Reorder terms.

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