Calculus Examples

$\frac{d y}{d x} = \frac{(x y + 3 x - y - 3)}{x y - 2 x + 4 y - 8}$

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor out the greatest common factor from each group.

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Step 1.1.1.1

Group the first two terms and the last two terms.

$\frac{d y}{d x} = \frac{(x y + 3 x) - y - 3}{x y - 2 x + 4 y - 8}$

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d y}{d x} = \frac{x (y + 3) - (y + 3)}{x y - 2 x + 4 y - 8}$

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $y + 3$ .

$\frac{d y}{d x} = \frac{(y + 3) (x - 1)}{x y - 2 x + 4 y - 8}$

Step 1.2

Factor.

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Step 1.2.1

Factor out the greatest common factor from each group.

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Step 1.2.1.1

Group the first two terms and the last two terms.

$\frac{d y}{d x} = \frac{(y + 3) (x - 1)}{(x y - 2 x) + 4 y - 8}$

Step 1.2.1.2

Factor out the greatest common factor (GCF) from each group.

$\frac{d y}{d x} = \frac{(y + 3) (x - 1)}{x (y - 2) + 4 (y - 2)}$

Step 1.2.2

Factor the polynomial by factoring out the greatest common factor, $y - 2$ .

$\frac{d y}{d x} = \frac{(y + 3) (x - 1)}{(y - 2) (x + 4)}$

Step 1.3

Regroup factors.

$\frac{d y}{d x} = \frac{x - 1}{x + 4} \cdot \frac{y + 3}{y - 2}$

Step 1.4

Multiply both sides by $\frac{y - 2}{y + 3}$ .

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{y - 2}{y + 3} (\frac{x - 1}{x + 4} \cdot \frac{y + 3}{y - 2})$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $\frac{x - 1}{x + 4}$ by $\frac{y + 3}{y - 2}$ .

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{y - 2}{y + 3} \cdot \frac{(x - 1) (y + 3)}{(x + 4) (y - 2)}$

Step 1.5.2

Cancel the common factor of $y - 2$ .

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Step 1.5.2.1

Factor $y - 2$ out of $(x + 4) (y - 2)$ .

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{y - 2}{y + 3} \cdot \frac{(x - 1) (y + 3)}{(y - 2) (x + 4)}$

Step 1.5.2.2

Cancel the common factor.

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{y - 2}{y + 3} \cdot \frac{(x - 1) (y + 3)}{(y - 2) (x + 4)}$

Step 1.5.2.3

Rewrite the expression.

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{1}{y + 3} \cdot \frac{(x - 1) (y + 3)}{x + 4}$

Step 1.5.3

Cancel the common factor of $y + 3$ .

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Step 1.5.3.1

Factor $y + 3$ out of $(x - 1) (y + 3)$ .

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{1}{y + 3} \cdot \frac{(y + 3) (x - 1)}{x + 4}$

Step 1.5.3.2

Cancel the common factor.

$\frac{y - 2}{y + 3} \frac{d y}{d x} = \frac{1}{y + 3} \cdot \frac{(y + 3) (x - 1)}{x + 4}$

Step 1.5.3.3

Rewrite the expression.

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

Step 1.6

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Divide $y - 2$ by $y + 3$ .

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Step 2.2.1.1

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

$y$

$3$

$y$

$2$

Step 2.2.1.2

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

					$1$
	$y$	+	$3$		$y$	-	$2$

Step 2.2.1.3

Multiply the new quotient term by the divisor.

				$1$
$y$	+	$3$		$y$	-	$2$
			+	$y$	+	$3$

Step 2.2.1.4

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

				$1$
$y$	+	$3$		$y$	-	$2$
			-	$y$	-	$3$

Step 2.2.1.5

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

				$1$
$y$	+	$3$		$y$	-	$2$
			-	$y$	-	$3$
					-	$5$

Step 2.2.1.6

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

$\int 1 - \frac{5}{y + 3} d y = \int \frac{x - 1}{x + 4} d x$

Step 2.2.2

Split the single integral into multiple integrals.

$\int d y + \int - \frac{5}{y + 3} d y = \int \frac{x - 1}{x + 4} d x$

Step 2.2.3

Apply the constant rule.

$y + C_{1} + \int - \frac{5}{y + 3} d y = \int \frac{x - 1}{x + 4} d x$

Step 2.2.4

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

$y + C_{1} - \int \frac{5}{y + 3} d y = \int \frac{x - 1}{x + 4} d x$

Step 2.2.5

Since $5$ is constant with respect to $y$ , move $5$ out of the integral.

$y + C_{1} - (5 \int \frac{1}{y + 3} d y) = \int \frac{x - 1}{x + 4} d x$

Step 2.2.6

Multiply $5$ by $- 1$ .

$y + C_{1} - 5 \int \frac{1}{y + 3} d y = \int \frac{x - 1}{x + 4} d x$

Step 2.2.7

Let $u_{1} = y + 3$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.7.1

Let $u_{1} = y + 3$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.7.1.1

Differentiate $y + 3$ .

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

Step 2.2.7.1.2

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

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

Step 2.2.7.1.3

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

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

Step 2.2.7.1.4

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

$1 + 0$

Step 2.2.7.1.5

Add $1$ and $0$ .

$1$

Step 2.2.7.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$y + C_{1} - 5 \int \frac{1}{u_{1}} d u_{1} = \int \frac{x - 1}{x + 4} d x$

Step 2.2.8

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

$y + C_{1} - 5 (\ln (| u_{1} |) + C_{2}) = \int \frac{x - 1}{x + 4} d x$

Step 2.2.9

Simplify.

$y - 5 \ln (| u_{1} |) + C_{3} = \int \frac{x - 1}{x + 4} d x$

Step 2.2.10

Replace all occurrences of $u_{1}$ with $y + 3$ .

$y - 5 \ln (| y + 3 |) + C_{3} = \int \frac{x - 1}{x + 4} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Divide $x - 1$ by $x + 4$ .

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Step 2.3.1.1

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

$x$

$4$

$x$

$1$

Step 2.3.1.2

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

					$1$
	$x$	+	$4$		$x$	-	$1$

Step 2.3.1.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$4$		$x$	-	$1$
			+	$x$	+	$4$

Step 2.3.1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 4$

				$1$
$x$	+	$4$		$x$	-	$1$
			-	$x$	-	$4$

Step 2.3.1.5

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

				$1$
$x$	+	$4$		$x$	-	$1$
			-	$x$	-	$4$
					-	$5$

Step 2.3.1.6

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

$y - 5 \ln (| y + 3 |) + C_{3} = \int 1 - \frac{5}{x + 4} d x$

Step 2.3.2

Split the single integral into multiple integrals.

$y - 5 \ln (| y + 3 |) + C_{3} = \int d x + \int - \frac{5}{x + 4} d x$

Step 2.3.3

Apply the constant rule.

$y - 5 \ln (| y + 3 |) + C_{3} = x + C_{4} + \int - \frac{5}{x + 4} d x$

Step 2.3.4

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

$y - 5 \ln (| y + 3 |) + C_{3} = x + C_{4} - \int \frac{5}{x + 4} d x$

Step 2.3.5

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$y - 5 \ln (| y + 3 |) + C_{3} = x + C_{4} - (5 \int \frac{1}{x + 4} d x)$

Step 2.3.6

Multiply $5$ by $- 1$ .

$y - 5 \ln (| y + 3 |) + C_{3} = x + C_{4} - 5 \int \frac{1}{x + 4} d x$

Step 2.3.7

Let $u_{2} = x + 4$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.7.1

Let $u_{2} = x + 4$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.7.1.1

Differentiate $x + 4$ .

$\frac{d}{d x} [x + 4]$

Step 2.3.7.1.2

By the Sum Rule, the derivative of $x + 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 2.3.7.1.3

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

$1 + \frac{d}{d x} [4]$

Step 2.3.7.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$1 + 0$

Step 2.3.7.1.5

Add $1$ and $0$ .

$1$

Step 2.3.7.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$y - 5 \ln (| y + 3 |) + C_{3} = x + C_{4} - 5 \int \frac{1}{u_{2}} d u_{2}$

Step 2.3.8

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$y - 5 \ln (| y + 3 |) + C_{3} = x + C_{4} - 5 (\ln (| u_{2} |) + C_{5})$

Step 2.3.9

Simplify.

$y - 5 \ln (| y + 3 |) + C_{3} = x - 5 \ln (| u_{2} |) + C_{6}$

Step 2.3.10

Replace all occurrences of $u_{2}$ with $x + 4$ .

$y - 5 \ln (| y + 3 |) + C_{3} = x - 5 \ln (| x + 4 |) + C_{6}$

Step 2.4

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

$y - 5 \ln (| y + 3 |) = x - 5 \ln (| x + 4 |) + C$