Calculus Examples

$\frac{6 x + 2}{3 x - 1}$

Step 1

Write $\frac{6 x + 2}{3 x - 1}$ as a function.

$f (x) = \frac{6 x + 2}{3 x - 1}$

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.

$F (x) = \int \frac{6 x + 2}{3 x - 1} d x$

Step 4

Divide $6 x + 2$ by $3 x - 1$ .

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Step 4.1

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

$3 x$

$1$

$6 x$

$2$

Step 4.2

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $3 x$ .

					$2$
	$3 x$	-	$1$		$6 x$	+	$2$

Step 4.3

Multiply the new quotient term by the divisor.

				$2$
$3 x$	-	$1$		$6 x$	+	$2$
			+	$6 x$	-	$2$

Step 4.4

The expression needs to be subtracted from the dividend, so change all the signs in $6 x - 2$

				$2$
$3 x$	-	$1$		$6 x$	+	$2$
			-	$6 x$	+	$2$

Step 4.5

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

				$2$
$3 x$	-	$1$		$6 x$	+	$2$
			-	$6 x$	+	$2$
					+	$4$

Step 4.6

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

$\int 2 + \frac{4}{3 x - 1} d x$

Step 5

Split the single integral into multiple integrals.

$\int 2 d x + \int \frac{4}{3 x - 1} d x$

Step 6

Apply the constant rule.

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

Step 7

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

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

Step 8

Let $u = 3 x - 1$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

Let $u = 3 x - 1$ . Find $\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate $3 x - 1$ .

$\frac{d}{d x} [3 x - 1]$

Step 8.1.2

By the Sum Rule, the derivative of $3 x - 1$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [3 x] + \frac{d}{d x} [- 1]$

Step 8.1.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 8.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 8.1.3.2

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

$3 \cdot 1 + \frac{d}{d x} [- 1]$

Step 8.1.3.3

Multiply $3$ by $1$ .

$3 + \frac{d}{d x} [- 1]$

Step 8.1.4

Differentiate using the Constant Rule.

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Step 8.1.4.1

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

$3 + 0$

Step 8.1.4.2

Add $3$ and $0$ .

$3$

Step 8.2

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

$2 x + C + 4 \int \frac{1}{u} \cdot \frac{1}{3} d u$

Step 9

Simplify.

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Step 9.1

Multiply $\frac{1}{u}$ by $\frac{1}{3}$ .

$2 x + C + 4 \int \frac{1}{u \cdot 3} d u$

Step 9.2

Move $3$ to the left of $u$ .

$2 x + C + 4 \int \frac{1}{3 u} d u$

Step 10

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

$2 x + C + 4 (\frac{1}{3} \int \frac{1}{u} d u)$

Step 11

Combine $\frac{1}{3}$ and $4$ .

$2 x + C + \frac{4}{3} \int \frac{1}{u} d u$

Step 12

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

$2 x + C + \frac{4}{3} (\ln (| u |) + C)$

Step 13

Simplify.

$2 x + \frac{4}{3} \ln (| u |) + C$

Step 14

Replace all occurrences of $u$ with $3 x - 1$ .

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

Step 15

The answer is the antiderivative of the function $f (x) = \frac{6 x + 2}{3 x - 1}$ .

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