Calculus Examples

$\int \frac{2 x^{2} - 7 x - 5}{2 x + 1} d x$

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 $d$ $u_{1}$ .

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Step 1.1

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

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

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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.

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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 $d u_{1}$ .

$\int \frac{2 {(\frac{u_{1}}{2} - \frac{1}{2})}^{2} - 7 (\frac{u_{1}}{2} - \frac{1}{2}) - 5}{u_{1}} \cdot \frac{1}{2} d u_{1}$

Step 2

Simplify.

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Step 2.1

Multiply $\frac{2 {(\frac{u_{1}}{2} - \frac{1}{2})}^{2} - 7 (\frac{u_{1}}{2} - \frac{1}{2}) - 5}{u_{1}}$ by $\frac{1}{2}$ .

$\int \frac{2 {(\frac{u_{1}}{2} - \frac{1}{2})}^{2} - 7 (\frac{u_{1}}{2} - \frac{1}{2}) - 5}{u_{1} \cdot 2} d u_{1}$

Step 2.2

Move $2$ to the left of $u_{1}$ .

$\int \frac{2 {(\frac{u_{1}}{2} - \frac{1}{2})}^{2} - 7 (\frac{u_{1}}{2} - \frac{1}{2}) - 5}{2 u_{1}} d u_{1}$

Step 3

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

$\frac{1}{2} \int \frac{2 {(\frac{u_{1}}{2} - \frac{1}{2})}^{2} - 7 (\frac{u_{1}}{2} - \frac{1}{2}) - 5}{u_{1}} d u_{1}$

Step 4

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 $d$ $u_{2}$ .

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

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 4.1.3

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

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Step 4.1.3.1

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 4.1.3.2

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 4.1.3.3

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

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

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

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 4.1.4.2

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

$\frac{1}{2}$

Step 4.2

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

$\frac{1}{2} \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} \cdot \frac{1}{\frac{1}{2}} d u_{2}$

Step 5

Simplify.

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Step 5.1

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

$\frac{1}{2} \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} (1 \cdot 2) d u_{2}$

Step 5.2

Multiply $2$ by $1$ .

$\frac{1}{2} \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} \cdot 2 d u_{2}$

Step 5.3

Combine $\frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1}$ and $2$ .

$\frac{1}{2} \int \frac{(2 {u_{2}}^{2} - 7 u_{2} - 5) \cdot 2}{2 u_{2} + 1} d u_{2}$

Step 5.4

Move $2$ to the left of $2 {u_{2}}^{2} - 7 u_{2} - 5$ .

$\frac{1}{2} \int \frac{2 (2 {u_{2}}^{2} - 7 u_{2} - 5)}{2 u_{2} + 1} d u_{2}$

Step 6

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

$\frac{1}{2} (2 \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} d u_{2})$

Step 7

Simplify.

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Step 7.1

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

$\frac{2}{2} \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} d u_{2}$

Step 7.2

Cancel the common factor of $2$ .

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Step 7.2.1

Cancel the common factor.

$\frac{2}{2} \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} d u_{2}$

Step 7.2.2

Rewrite the expression.

$1 \int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} d u_{2}$

Step 7.3

Multiply $\int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} d u_{2}$ by $1$ .

$\int \frac{2 {u_{2}}^{2} - 7 u_{2} - 5}{2 u_{2} + 1} d u_{2}$

Step 8

Divide $2 {u_{2}}^{2} - 7 u_{2} - 5$ by $2 u_{2} + 1$ .

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Step 8.1

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

$2 u_{2}$

$1$

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

$7 u_{2}$

$5$

Step 8.2

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

					$u_{2}$
	$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$

Step 8.3

Multiply the new quotient term by the divisor.

				$u_{2}$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			+	$2 {u_{2}}^{2}$	+	$u_{2}$

Step 8.4

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

				$u_{2}$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$

Step 8.5

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

				$u_{2}$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$
					-	$8 u_{2}$

Step 8.6

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

				$u_{2}$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$
					-	$8 u_{2}$	-	$5$

Step 8.7

Divide the highest order term in the dividend $- 8 u_{2}$ by the highest order term in divisor $2 u_{2}$ .

				$u_{2}$	-	$4$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$
					-	$8 u_{2}$	-	$5$

Step 8.8

Multiply the new quotient term by the divisor.

				$u_{2}$	-	$4$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$
					-	$8 u_{2}$	-	$5$
					-	$8 u_{2}$	-	$4$

Step 8.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 u_{2} - 4$

				$u_{2}$	-	$4$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$
					-	$8 u_{2}$	-	$5$
					+	$8 u_{2}$	+	$4$

Step 8.10

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

				$u_{2}$	-	$4$
$2 u_{2}$	+	$1$		$2 {u_{2}}^{2}$	-	$7 u_{2}$	-	$5$
			-	$2 {u_{2}}^{2}$	-	$u_{2}$
					-	$8 u_{2}$	-	$5$
					+	$8 u_{2}$	+	$4$
							-	$1$

Step 8.11

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

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

Step 9

Split the single integral into multiple integrals.

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

Step 10

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

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

Step 11

Apply the constant rule.

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

Step 12

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

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

Step 13

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 $d$ $u_{3}$ .

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Step 13.1

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

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Step 13.1.1

Differentiate $2 u_{2} + 1$ .

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

Step 13.1.2

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 13.1.3

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

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Step 13.1.3.1

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 13.1.3.2

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 13.1.3.3

Multiply $2$ by $1$ .

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

Step 13.1.4

Differentiate using the Constant Rule.

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Step 13.1.4.1

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

$2 + 0$

Step 13.1.4.2

Add $2$ and $0$ .

$2$

Step 13.2

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

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

Step 14

Simplify.

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Step 14.1

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

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

Step 14.2

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

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

Step 15

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

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

Step 16

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

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

Step 17

Simplify.

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

Step 18

Substitute back in for each integration substitution variable.

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Step 18.1

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

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

Step 18.2

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

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

Step 18.3

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

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

Step 18.4

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

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

Step 18.5

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

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

Step 19

Simplify.

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Step 19.1

Combine the numerators over the common denominator.

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

Step 19.2

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

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Step 19.2.1

Subtract $1$ from $1$ .

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

Step 19.2.2

Add $2 x$ and $0$ .

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

Step 19.3

Cancel the common factor of $2$ .

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Step 19.3.1

Factor $2$ out of $- 4$ .

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

Step 19.3.2

Cancel the common factor.

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

Step 19.3.3

Rewrite the expression.

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

Step 19.4

Multiply $2$ by $- 2$ .

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

Step 19.5

Combine the numerators over the common denominator.

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

Step 19.6

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

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Step 19.6.1

Subtract $1$ from $1$ .

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

Step 19.6.2

Add $2 x$ and $0$ .

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

Step 19.7

Cancel the common factor of $2$ .

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Step 19.7.1

Cancel the common factor.

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

Step 19.7.2

Rewrite the expression.

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

Step 19.8

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

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

Step 20

Reorder terms.

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