Calculus Examples

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

Step 1

Divide $x + 3$ by $4 x + 4$ .

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Step 1.1

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

$4 x$

$4$

$x$

$3$

Step 1.2

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

					$\frac{1}{4}$
	$4 x$	+	$4$		$x$	+	$3$

Step 1.3

Multiply the new quotient term by the divisor.

				$\frac{1}{4}$
$4 x$	+	$4$		$x$	+	$3$
			+	$x$	+	$1$

Step 1.4

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

				$\frac{1}{4}$
$4 x$	+	$4$		$x$	+	$3$
			-	$x$	-	$1$

Step 1.5

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

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

Step 1.6

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

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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Step 3.1

Factor $2$ out of $2$ .

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

Step 3.2

Cancel the common factors.

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Step 3.2.1

Factor $2$ out of $4 x$ .

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

Step 3.2.2

Factor $2$ out of $4$ .

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

Step 3.2.3

Factor $2$ out of $2 (2 x) + 2 (2)$ .

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

Step 3.2.4

Cancel the common factor.

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

Step 3.2.5

Rewrite the expression.

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

Step 4

Apply the constant rule.

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

Step 5

Let $u = 2 x + 2$ . 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 5.1

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

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Step 5.1.1

Differentiate $2 x + 2$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

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Step 5.1.3.1

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

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

Step 5.1.3.2

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

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

Step 5.1.3.3

Multiply $2$ by $1$ .

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

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$2 + 0$

Step 5.1.4.2

Add $2$ and $0$ .

$2$

Step 5.2

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

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

Step 6

Simplify.

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Step 6.1

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

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

Step 6.2

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

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

Step 7

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

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

Step 8

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

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

Step 9

Simplify.

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

Step 10

Replace all occurrences of $u$ with $2 x + 2$ .

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