Calculus Examples

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

Step 1

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

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

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.

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

Step 4

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

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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 $0$ .

$2 x$

$3$

$2 x$

$1$

Step 4.2

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

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

Step 4.3

Multiply the new quotient term by the divisor.

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

Step 4.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x - 3$

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

Step 4.5

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

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

Step 4.6

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

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

Apply the constant rule.

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

Step 7

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

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

Step 8

Let $u = 2 x - 3$ . 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 8.1

Let $u = 2 x - 3$ . Find $\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate $2 x - 3$ .

$\frac{d}{d x} [2 x - 3]$

Step 8.1.2

By the Sum Rule, the derivative of $2 x - 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$

Step 8.1.3

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

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Step 8.1.3.1

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

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

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 $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [- 3]$

Step 8.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 3]$

Step 8.1.4

Differentiate using the Constant Rule.

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Step 8.1.4.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$2 + 0$

Step 8.1.4.2

Add $2$ and $0$ .

$2$

Step 8.2

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

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

Step 9

Simplify.

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Step 9.1

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

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

Step 9.2

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

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

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

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

Step 11.2

Cancel the common factor of $4$ and $2$ .

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Step 11.2.1

Factor $2$ out of $4$ .

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

Step 11.2.2

Cancel the common factors.

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Step 11.2.2.1

Factor $2$ out of $2$ .

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

Step 11.2.2.2

Cancel the common factor.

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

Step 11.2.2.3

Rewrite the expression.

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

Step 11.2.2.4

Divide $2$ by $1$ .

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

Step 12

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

$x + C + 2 (\ln (| u |) + C)$

Step 13

Simplify.

$x + 2 \ln (| u |) + C$

Step 14

Replace all occurrences of $u$ with $2 x - 3$ .

$x + 2 \ln (| 2 x - 3 |) + C$

Step 15

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

$F (x) =$ $x + 2 \ln (| 2 x - 3 |) + C$