Calculus Examples

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

Step 1

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

Tap for more steps...

Step 1.1

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

Tap for more steps...

Step 1.1.1

Differentiate $2 x + 1$ .

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

Step 1.1.2

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 1.1.3

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

Tap for more steps...

Step 1.1.3.1

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 1.1.3.2

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 1.1.3.3

Multiply $2$ by $1$ .

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

Step 1.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 1.1.4.1

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

$2 + 0$

Step 1.1.4.2

Add $2$ and $0$ .

$2$

Step 1.2

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

$\int \frac{6 {(\frac{u_{1}}{2} - \frac{1}{2})}^{2}}{u_{1}} d u_{1}$

Step 2

Since $6$ is constant with respect to $u_{1}$ , move $6$ out of the integral.

$6 \int \frac{{(\frac{u_{1}}{2} - \frac{1}{2})}^{2}}{u_{1}} d u_{1}$

Step 3

Let $u_{2} = \frac{u_{1}}{2} - \frac{1}{2}$ . Then $d u_{2} = \frac{1}{2} d u_{1}$ , so $2 d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

Differentiate $\frac{u_{1}}{2} - \frac{1}{2}$ .

$\frac{d}{d u_{1}} [\frac{u_{1}}{2} - \frac{1}{2}]$

Step 3.1.2

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

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

Step 3.1.3

Evaluate $\frac{d}{d u_{1}} [\frac{u_{1}}{2}]$ .

Tap for more steps...

Step 3.1.3.1

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , the derivative of $\frac{u_{1}}{2}$ with respect to $u_{1}$ is $\frac{1}{2} \frac{d}{d u_{1}} [u_{1}]$ .

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

Step 3.1.3.2

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

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

Step 3.1.3.3

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

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

Step 3.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 3.1.4.1

Since $- \frac{1}{2}$ is constant with respect to $u_{1}$ , the derivative of $- \frac{1}{2}$ with respect to $u_{1}$ is $0$ .

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

Step 3.1.4.2

Add $\frac{1}{2}$ and $0$ .

$\frac{1}{2}$

Step 3.2

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

$6 \int \frac{{u_{2}}^{2}}{2 u_{2} + 1} \cdot \frac{1}{\frac{1}{2}} d u_{2}$

Step 4

Simplify.

Tap for more steps...

Step 4.1

Multiply by the reciprocal of the fraction to divide by $\frac{1}{2}$ .

$6 \int \frac{{u_{2}}^{2}}{2 u_{2} + 1} (1 \cdot 2) d u_{2}$

Step 4.2

Multiply $2$ by $1$ .

$6 \int \frac{{u_{2}}^{2}}{2 u_{2} + 1} \cdot 2 d u_{2}$

Step 4.3

Combine $\frac{{u_{2}}^{2}}{2 u_{2} + 1}$ and $2$ .

$6 \int \frac{{u_{2}}^{2} \cdot 2}{2 u_{2} + 1} d u_{2}$

Step 4.4

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

$6 \int \frac{2 {u_{2}}^{2}}{2 u_{2} + 1} d u_{2}$

Step 5

Since $2$ is constant with respect to $u_{2}$ , move $2$ out of the integral.

$6 (2 \int \frac{{u_{2}}^{2}}{2 u_{2} + 1} d u_{2})$

Step 6

Multiply $2$ by $6$ .

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

Step 7

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

Tap for more steps...

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 $0$ .

$2 u_{2}$

$1$

${u_{2}}^{2}$

$0 u_{2}$

$0$

Step 7.2

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

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

Step 7.3

Multiply the new quotient term by the divisor.

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

Step 7.4

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

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

Step 7.5

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

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

Step 7.6

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

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

Step 7.7

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

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

Step 7.8

Multiply the new quotient term by the divisor.

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

Step 7.9

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

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

Step 7.10

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

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

Step 7.11

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

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

Step 8

Split the single integral into multiple integrals.

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

Step 9

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

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

Step 10

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

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

Step 11

Apply the constant rule.

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

Step 12

Simplify.

Tap for more steps...

Step 12.1

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

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

Step 12.2

Combine $u_{2}$ and $\frac{1}{4}$ .

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

Step 13

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

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

Step 14

Let $u_{3} = 2 u_{2} + 1$ . Then $d u_{3} = 2 d u_{2}$ , so $\frac{1}{2} d u_{3} = d u_{2}$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

Tap for more steps...

Step 14.1

Let $u_{3} = 2 u_{2} + 1$ . Find $\frac{d u_{3}}{d u_{2}}$ .

Tap for more steps...

Step 14.1.1

Differentiate $2 u_{2} + 1$ .

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

Step 14.1.2

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

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

Step 14.1.3

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

Tap for more steps...

Step 14.1.3.1

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

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

Step 14.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 1$ .

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

Step 14.1.3.3

Multiply $2$ by $1$ .

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

Step 14.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 14.1.4.1

Since $1$ is constant with respect to $u_{2}$ , the derivative of $1$ with respect to $u_{2}$ is $0$ .

$2 + 0$

Step 14.1.4.2

Add $2$ and $0$ .

$2$

Step 14.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$12 (\frac{1}{2} (\frac{{u_{2}}^{2}}{2} + C) - \frac{u_{2}}{4} + C + \frac{1}{4} \int \frac{1}{u_{3}} \cdot \frac{1}{2} d u_{3})$

Step 15

Simplify.

Tap for more steps...

Step 15.1

Multiply $\frac{1}{u_{3}}$ by $\frac{1}{2}$ .

$12 (\frac{1}{2} (\frac{{u_{2}}^{2}}{2} + C) - \frac{u_{2}}{4} + C + \frac{1}{4} \int \frac{1}{u_{3} \cdot 2} d u_{3})$

Step 15.2

Move $2$ to the left of $u_{3}$ .

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

Step 16

Since $\frac{1}{2}$ is constant with respect to $u_{3}$ , move $\frac{1}{2}$ out of the integral.

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

Step 17

Simplify.

Tap for more steps...

Step 17.1

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

$12 (\frac{1}{2} (\frac{{u_{2}}^{2}}{2} + C) - \frac{u_{2}}{4} + C + \frac{1}{2 \cdot 4} \int \frac{1}{u_{3}} d u_{3})$

Step 17.2

Multiply $2$ by $4$ .

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

Step 18

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

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

Step 19

Simplify.

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

Step 20

Reorder terms.

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

Step 21

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 21.1

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

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

Step 21.2

Replace all occurrences of $u_{3}$ with $2 u_{2} + 1$ .

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

Step 21.3

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

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

Step 21.4

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

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

Step 21.5

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

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

Step 22

Simplify.

Tap for more steps...

Step 22.1

Combine the numerators over the common denominator.

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

Step 22.2

Combine the opposite terms in $2 x + 1 - 1$ .

Tap for more steps...

Step 22.2.1

Subtract $1$ from $1$ .

$12 (\frac{1}{4} {(\frac{2 x + 1}{2} - \frac{1}{2})}^{2} - \frac{1}{4} \cdot \frac{2 x + 0}{2} + \frac{1}{8} \ln (| 2 (\frac{2 x + 1}{2} - \frac{1}{2}) + 1 |)) + C$

Step 22.2.2

Add $2 x$ and $0$ .

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

Step 22.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 22.3.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

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

Step 22.3.2

Factor $2$ out of $4$ .

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

Step 22.3.3

Factor $2$ out of $2 x$ .

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

Step 22.3.4

Cancel the common factor.

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

Step 22.3.5

Rewrite the expression.

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

Step 22.4

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

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

Step 22.5

Multiply $2$ by $2$ .

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

Step 22.6

Move the negative in front of the fraction.

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

Step 22.7

Combine the numerators over the common denominator.

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

Step 22.8

Combine the opposite terms in $2 x + 1 - 1$ .

Tap for more steps...

Step 22.8.1

Subtract $1$ from $1$ .

$12 (\frac{1}{4} {(\frac{2 x + 1}{2} - \frac{1}{2})}^{2} - \frac{x}{4} + \frac{1}{8} \ln (| 2 \frac{2 x + 0}{2} + 1 |)) + C$

Step 22.8.2

Add $2 x$ and $0$ .

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

Step 22.9

Cancel the common factor of $2$ .

Tap for more steps...

Step 22.9.1

Cancel the common factor.

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

Step 22.9.2

Rewrite the expression.

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

Step 22.10

Simplify each term.

Tap for more steps...

Step 22.10.1

Combine the numerators over the common denominator.

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

Step 22.10.2

Combine the opposite terms in $2 x + 1 - 1$ .

Tap for more steps...

Step 22.10.2.1

Subtract $1$ from $1$ .

$12 (\frac{1}{4} {(\frac{2 x + 0}{2})}^{2} - \frac{x}{4} + \frac{1}{8} \ln (| 2 x + 1 |)) + C$

Step 22.10.2.2

Add $2 x$ and $0$ .

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

Step 22.10.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 22.10.3.1

Cancel the common factor.

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

Step 22.10.3.2

Divide $x$ by $1$ .

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

Step 22.10.4

Combine $\frac{1}{4}$ and $x^{2}$ .

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

Step 22.10.5

Combine $\frac{1}{8}$ and $\ln (| 2 x + 1 |)$ .

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

Step 22.11

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 22.12

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

Tap for more steps...

Step 22.12.1

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 22.12.2

Multiply $4$ by $2$ .

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

Step 22.13

Combine the numerators over the common denominator.

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

Step 22.14

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

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

Step 22.15

Apply the distributive property.

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

Step 22.16

Cancel the common factor of $4$ .

Tap for more steps...

Step 22.16.1

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 22.16.2

Factor $4$ out of $12$ .

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

Step 22.16.3

Cancel the common factor.

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

Step 22.16.4

Rewrite the expression.

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

Step 22.17

Multiply $- 1$ by $3$ .

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

Step 22.18

Cancel the common factor of $4$ .

Tap for more steps...

Step 22.18.1

Factor $4$ out of $12$ .

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

Step 22.18.2

Factor $4$ out of $8$ .

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

Step 22.18.3

Cancel the common factor.

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

Step 22.18.4

Rewrite the expression.

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

Step 22.19

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 22.20

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 22.21

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 22.22

Combine the numerators over the common denominator.

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

Step 22.23

Simplify the numerator.

Tap for more steps...

Step 22.23.1

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

Tap for more steps...

Step 22.23.1.1

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

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

Step 22.23.1.2

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 22.23.2

Multiply $2$ by $- 1$ .

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

Step 22.24

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

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

Step 22.25

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

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

Step 22.26

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

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

Step 22.27

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

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

Step 22.28

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 22.29

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 22.30

Move the negative in front of the fraction.

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

Step 23

Reorder terms.

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