Calculus Examples

$\int \ln (4 x + 1) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (4 x + 1)$ and $d v = 1$ .

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

Step 2

Simplify.

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Step 2.1

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

$\ln (4 x + 1) x - \int \frac{x \cdot 4}{4 x + 1} d x$

Step 2.2

Move $4$ to the left of $x$ .

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

Step 3

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

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

Step 4

Multiply $4$ by $- 1$ .

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

Step 5

Divide $x$ by $4 x + 1$ .

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Step 5.1

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

$4 x$

$1$

$x$

$0$

Step 5.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $4 x$ .

					$\frac{1}{4}$
	$4 x$	+	$1$		$x$	+	$0$

Step 5.3

Multiply the new quotient term by the divisor.

				$\frac{1}{4}$
$4 x$	+	$1$		$x$	+	$0$
			+	$x$	+	$\frac{1}{4}$

Step 5.4

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

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

Step 5.5

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

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

Step 5.6

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

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

Apply the constant rule.

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

Let $u = 4 x + 1$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 11.1

Let $u = 4 x + 1$ . Find $\frac{d u}{d x}$ .

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Step 11.1.1

Differentiate $4 x + 1$ .

$\frac{d}{d x} [4 x + 1]$

Step 11.1.2

By the Sum Rule, the derivative of $4 x + 1$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [1]$

Step 11.1.3

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 11.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 11.1.3.2

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

$4 \cdot 1 + \frac{d}{d x} [1]$

Step 11.1.3.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [1]$

Step 11.1.4

Differentiate using the Constant Rule.

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Step 11.1.4.1

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

$4 + 0$

Step 11.1.4.2

Add $4$ and $0$ .

$4$

Step 11.2

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

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

Step 12

Simplify.

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Step 12.1

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

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

Step 12.2

Move $4$ to the left of $u$ .

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

Step 13

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

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

Step 14

Simplify.

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Step 14.1

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

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

Step 14.2

Multiply $4$ by $4$ .

$\ln (4 x + 1) x - 4 (\frac{x}{4} + C - \frac{1}{16} \int \frac{1}{u} d u)$

Step 15

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

$\ln (4 x + 1) x - 4 (\frac{x}{4} + C - \frac{1}{16} (\ln (| u |) + C))$

Step 16

Simplify.

$\ln (4 x + 1) x - 4 (\frac{1}{4} x - \frac{1}{16} \ln (| u |)) + C$

Step 17

Replace all occurrences of $u$ with $4 x + 1$ .

$\ln (4 x + 1) x - 4 (\frac{1}{4} x - \frac{1}{16} \ln (| 4 x + 1 |)) + C$

Step 18

Simplify.

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Step 18.1

Simplify each term.

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Step 18.1.1

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

$\ln (4 x + 1) x - 4 (\frac{x}{4} - \frac{1}{16} \ln (| 4 x + 1 |)) + C$

Step 18.1.2

Combine $\ln (| 4 x + 1 |)$ and $\frac{1}{16}$ .

$\ln (4 x + 1) x - 4 (\frac{x}{4} - \frac{\ln (| 4 x + 1 |)}{16}) + C$

Step 18.2

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

$\ln (4 x + 1) x - 4 (\frac{x}{4} \cdot \frac{4}{4} - \frac{\ln (| 4 x + 1 |)}{16}) + C$

Step 18.3

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Step 18.3.1

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

$\ln (4 x + 1) x - 4 (\frac{x \cdot 4}{4 \cdot 4} - \frac{\ln (| 4 x + 1 |)}{16}) + C$

Step 18.3.2

Multiply $4$ by $4$ .

$\ln (4 x + 1) x - 4 (\frac{x \cdot 4}{16} - \frac{\ln (| 4 x + 1 |)}{16}) + C$

Step 18.4

Combine the numerators over the common denominator.

$\ln (4 x + 1) x - 4 \frac{x \cdot 4 - \ln (| 4 x + 1 |)}{16} + C$

Step 18.5

Cancel the common factor of $4$ .

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Step 18.5.1

Factor $4$ out of $- 4$ .

$\ln (4 x + 1) x + 4 (- 1) \frac{x \cdot 4 - \ln (| 4 x + 1 |)}{16} + C$

Step 18.5.2

Factor $4$ out of $16$ .

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

Step 18.5.3

Cancel the common factor.

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

Step 18.5.4

Rewrite the expression.

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

Step 18.6

Move $4$ to the left of $x$ .

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

Step 19

Reorder terms.

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