Calculus Examples

$\frac{d z}{d x} = \frac{z + 4}{2 z - 1} + 1$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{\frac{z + 4}{2 z - 1} + 1}$ .

$\frac{1}{\frac{z + 4}{2 z - 1} + 1} \frac{d z}{d x} = \frac{1}{\frac{z + 4}{2 z - 1} + 1} (\frac{z + 4}{2 z - 1} + 1)$

Step 1.2

Cancel the common factor of $\frac{z + 4}{2 z - 1} + 1$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{\frac{z + 4}{2 z - 1} + 1} \frac{d z}{d x} = \frac{1}{\frac{z + 4}{2 z - 1} + 1} (\frac{z + 4}{2 z - 1} + 1)$

Step 1.2.2

Rewrite the expression.

$\frac{1}{\frac{z + 4}{2 z - 1} + 1} \frac{d z}{d x} = 1$

Step 1.3

Rewrite the equation.

$\frac{1}{\frac{z + 4}{2 z - 1} + 1} d z = 1 d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify.

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Step 2.2.1.1

Write $1$ as a fraction with a common denominator.

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

Step 2.2.1.2

Combine the numerators over the common denominator.

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

Step 2.2.1.3

Add $z$ and $2 z$ .

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

Step 2.2.1.4

Subtract $1$ from $4$ .

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

Step 2.2.1.5

Multiply by the reciprocal of the fraction to divide by $\frac{3 z + 3}{2 z - 1}$ .

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

Step 2.2.1.6

Multiply $\frac{2 z - 1}{3 z + 3}$ by $1$ .

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

Step 2.2.2

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

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Step 2.2.2.1

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

$3 z$

$3$

$2 z$

$1$

Step 2.2.2.2

Divide the highest order term in the dividend $2 z$ by the highest order term in divisor $3 z$ .

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

Step 2.2.2.3

Multiply the new quotient term by the divisor.

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

Step 2.2.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 z + 2$

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

Step 2.2.2.5

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

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

Step 2.2.2.6

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

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

Step 2.2.3

Split the single integral into multiple integrals.

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

Step 2.2.4

Cancel the common factor of $3$ and $3 z + 3$ .

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Step 2.2.4.1

Factor $3$ out of $3$ .

$\int \frac{2}{3} d z + \int - \frac{3 \cdot 1}{3 z + 3} d z = \int d x$

Step 2.2.4.2

Cancel the common factors.

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Step 2.2.4.2.1

Factor $3$ out of $3 z$ .

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

Step 2.2.4.2.2

Factor $3$ out of $3$ .

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

Step 2.2.4.2.3

Factor $3$ out of $3 (z) + 3 (1)$ .

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

Step 2.2.4.2.4

Cancel the common factor.

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

Step 2.2.4.2.5

Rewrite the expression.

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

Step 2.2.5

Apply the constant rule.

$\frac{2}{3} z + C_{1} + \int - \frac{1}{z + 1} d z = \int d x$

Step 2.2.6

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

$\frac{2}{3} z + C_{1} - \int \frac{1}{z + 1} d z = \int d x$

Step 2.2.7

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

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Step 2.2.7.1

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

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Step 2.2.7.1.1

Differentiate $z + 1$ .

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

Step 2.2.7.1.2

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

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

Step 2.2.7.1.3

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

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

Step 2.2.7.1.4

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

$1 + 0$

Step 2.2.7.1.5

Add $1$ and $0$ .

$1$

Step 2.2.7.2

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

$\frac{2}{3} z + C_{1} - \int \frac{1}{u} d u = \int d x$

Step 2.2.8

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

$\frac{2}{3} z + C_{1} - (\ln (| u |) + C_{2}) = \int d x$

Step 2.2.9

Simplify.

$\frac{2}{3} z - \ln (| u |) + C_{3} = \int d x$

Step 2.2.10

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

$\frac{2}{3} z - \ln (| z + 1 |) + C_{3} = \int d x$

Step 2.3

Apply the constant rule.

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

Step 2.4

Group the constant of integration on the right side as $C$ .

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