Calculus Examples

$\int (\frac{x - 3}{x + 3}) d x$

Step 1

Remove parentheses.

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

Step 2

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

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Step 2.1

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

$x$

$3$

$x$

$3$

Step 2.2

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

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

Step 2.3

Multiply the new quotient term by the divisor.

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

$\int 1 - \frac{6}{x + 3} d x$

Step 3

Split the single integral into multiple integrals.

$\int d x + \int - \frac{6}{x + 3} d x$

Step 4

Apply the constant rule.

$x + C + \int - \frac{6}{x + 3} d x$

Step 5

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

$x + C - \int \frac{6}{x + 3} d x$

Step 6

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$x + C - (6 \int \frac{1}{x + 3} d x)$

Step 7

Multiply $6$ by $- 1$ .

$x + C - 6 \int \frac{1}{x + 3} d x$

Step 8

Let $u = x + 3$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

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

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Step 8.1.1

Differentiate $x + 3$ .

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

Step 8.1.2

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

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

Step 8.1.3

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

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

Step 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

$x + C - 6 \int \frac{1}{u} d u$

Step 9

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

$x + C - 6 (\ln (| u |) + C)$

Step 10

Simplify.

$x - 6 \ln (| u |) + C$

Step 11

Replace all occurrences of $u$ with $x + 3$ .

$x - 6 \ln (| x + 3 |) + C$