Calculus Examples

$\int \frac{1}{2 + 2 w} \cdot (2 w) d w$

Step 1

Simplify.

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Step 1.1

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

$\int \frac{2}{2 + 2 w} \cdot w d w$

Step 1.2

Combine $\frac{2}{2 + 2 w}$ and $w$ .

$\int \frac{2 w}{2 + 2 w} d w$

Step 1.3

Cancel the common factor of $2$ and $2 + 2 w$ .

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Step 1.3.1

Factor $2$ out of $2 w$ .

$\int \frac{2 (w)}{2 + 2 w} d w$

Step 1.3.2

Cancel the common factors.

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Step 1.3.2.1

Factor $2$ out of $2$ .

$\int \frac{2 (w)}{2 \cdot 1 + 2 w} d w$

Step 1.3.2.2

Factor $2$ out of $2 w$ .

$\int \frac{2 (w)}{2 \cdot 1 + 2 (w)} d w$

Step 1.3.2.3

Factor $2$ out of $2 \cdot 1 + 2 (w)$ .

$\int \frac{2 (w)}{2 \cdot (1 + w)} d w$

Step 1.3.2.4

Cancel the common factor.

$\int \frac{2 w}{2 \cdot (1 + w)} d w$

Step 1.3.2.5

Rewrite the expression.

$\int \frac{w}{1 + w} d w$

Step 2

Reorder $1$ and $w$ .

$\int \frac{w}{w + 1} d w$

Step 3

Divide $w$ by $w + 1$ .

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Step 3.1

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

$w$

$1$

$w$

$0$

Step 3.2

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

					$1$
	$w$	+	$1$		$w$	+	$0$

Step 3.3

Multiply the new quotient term by the divisor.

				$1$
$w$	+	$1$		$w$	+	$0$
			+	$w$	+	$1$

Step 3.4

The expression needs to be subtracted from the dividend, so change all the signs in $w + 1$

				$1$
$w$	+	$1$		$w$	+	$0$
			-	$w$	-	$1$

Step 3.5

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

				$1$
$w$	+	$1$		$w$	+	$0$
			-	$w$	-	$1$
					-	$1$

Step 3.6

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

$\int 1 - \frac{1}{w + 1} d w$

Step 4

Split the single integral into multiple integrals.

$\int d w + \int - \frac{1}{w + 1} d w$

Step 5

Apply the constant rule.

$w + C + \int - \frac{1}{w + 1} d w$

Step 6

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

$w + C - \int \frac{1}{w + 1} d w$

Step 7

Let $u = w + 1$ . Then $d u = d w$ . Rewrite using $u$ and $d$ $u$ .

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Step 7.1

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

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Step 7.1.1

Differentiate $w + 1$ .

$\frac{d}{d w} [w + 1]$

Step 7.1.2

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

$\frac{d}{d w} [w] + \frac{d}{d w} [1]$

Step 7.1.3

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

$1 + \frac{d}{d w} [1]$

Step 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

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

Step 8

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

$w + C - (\ln (| u |) + C)$

Step 9

Simplify.

$w - \ln (| u |) + C$

Step 10

Replace all occurrences of $u$ with $w + 1$ .

$w - \ln (| w + 1 |) + C$