Calculus Examples

$\ln (2 x + 1)$

Step 1

Write $\ln (2 x + 1)$ as a function.

$f (x) = \ln (2 x + 1)$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \ln (2 x + 1) d x$

Step 4

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (2 x + 1)$ and $d v = 1$ .

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

Step 5

Simplify.

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Step 5.1

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

$\ln (2 x + 1) x - \int \frac{x \cdot 2}{2 x + 1} d x$

Step 5.2

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

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

Step 6

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

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

Step 7

Multiply $2$ by $- 1$ .

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

Step 8

Divide $x$ by $2 x + 1$ .

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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 $0$ .

$2 x$

$1$

$x$

$0$

Step 8.2

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

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

Step 8.3

Multiply the new quotient term by the divisor.

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

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

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

Step 9

Split the single integral into multiple integrals.

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

Step 10

Apply the constant rule.

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

Step 11

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

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

Step 12

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

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

Step 13

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

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

Step 14

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 14.1

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

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Step 14.1.1

Differentiate $2 x + 1$ .

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

Step 14.1.2

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 14.1.3

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

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Step 14.1.3.1

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 14.1.3.2

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 14.1.3.3

Multiply $2$ by $1$ .

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

Step 14.1.4

Differentiate using the Constant Rule.

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Step 14.1.4.1

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

$2 + 0$

Step 14.1.4.2

Add $2$ and $0$ .

$2$

Step 14.2

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

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

Step 15

Simplify.

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Step 15.1

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

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

Step 15.2

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

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

Step 16

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

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

Step 17

Simplify.

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Step 17.1

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

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

Step 17.2

Multiply $2$ by $2$ .

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

Step 18

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

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

Step 19

Simplify.

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

Step 20

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

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

Step 21

Simplify.

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Step 21.1

Simplify each term.

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Step 21.1.1

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

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

Step 21.1.2

Combine $\ln (| 2 x + 1 |)$ and $\frac{1}{4}$ .

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

Step 21.2

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

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

Step 21.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 21.3.1

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

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

Step 21.3.2

Multiply $2$ by $2$ .

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

Step 21.4

Combine the numerators over the common denominator.

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

Step 21.5

Cancel the common factor of $2$ .

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Step 21.5.1

Factor $2$ out of $- 2$ .

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

Step 21.5.2

Factor $2$ out of $4$ .

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

Step 21.5.3

Cancel the common factor.

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

Step 21.5.4

Rewrite the expression.

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

Step 21.6

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

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

Step 22

Reorder terms.

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

Step 23

The answer is the antiderivative of the function $f (x) = \ln (2 x + 1)$ .

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