Calculus Examples

$x (y^{2} - 1) d y - y (x^{2} - 1) d x = 0$

Step 1

Add $y (x^{2} - 1) d x$ to both sides of the equation.

$x (y^{2} - 1) d y = y (x^{2} - 1) d x$

Step 2

Multiply both sides by $\frac{1}{x y}$ .

$\frac{1}{x y} (x (y^{2} - 1)) d y = \frac{1}{x y} (y (x^{2} - 1)) d x$

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Factor $x$ out of $x y$ .

$\frac{1}{x (y)} (x (y^{2} - 1)) d y = \frac{1}{x y} (y (x^{2} - 1)) d x$

Step 3.1.2

Cancel the common factor.

$\frac{1}{x y} (x (y^{2} - 1)) d y = \frac{1}{x y} (y (x^{2} - 1)) d x$

Step 3.1.3

Rewrite the expression.

$\frac{1}{y} (y^{2} - 1) d y = \frac{1}{x y} (y (x^{2} - 1)) d x$

Step 3.2

Multiply $\frac{1}{y}$ by $y^{2} - 1$ .

$\frac{y^{2} - 1}{y} d y = \frac{1}{x y} (y (x^{2} - 1)) d x$

Step 3.3

Simplify the numerator.

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Step 3.3.1

Rewrite $1$ as $1^{2}$ .

$\frac{y^{2} - 1^{2}}{y} d y = \frac{1}{x y} (y (x^{2} - 1)) d x$

Step 3.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 1$ .

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

Step 3.4

Cancel the common factor of $y$ .

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Step 3.4.1

Factor $y$ out of $x y$ .

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

Step 3.4.2

Cancel the common factor.

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

Step 3.4.3

Rewrite the expression.

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

Step 3.5

Multiply $\frac{1}{x}$ by $x^{2} - 1$ .

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

Step 3.6

Simplify the numerator.

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Step 3.6.1

Rewrite $1$ as $1^{2}$ .

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

Step 3.6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{(y + 1) (y - 1)}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Apply the distributive property.

$\int \frac{y (y - 1) + 1 (y - 1)}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.2

Apply the distributive property.

$\int \frac{y \cdot y + y \cdot - 1 + 1 (y - 1)}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.3

Apply the distributive property.

$\int \frac{y \cdot y + y \cdot - 1 + 1 y + 1 \cdot - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.4

Reorder $y$ and $- 1$ .

$\int \frac{y \cdot y - 1 \cdot y + 1 y + 1 \cdot - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.5

Raise $y$ to the power of $1$ .

$\int \frac{y^{1} y - 1 y + 1 y + 1 \cdot - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.6

Raise $y$ to the power of $1$ .

$\int \frac{y^{1} y^{1} - 1 y + 1 y + 1 \cdot - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int \frac{y^{1 + 1} - 1 y + 1 y + 1 \cdot - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.8

Simplify the expression.

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Step 4.2.8.1

Add $1$ and $1$ .

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

Step 4.2.8.2

Multiply $y$ by $1$ .

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

Step 4.2.8.3

Multiply $- 1$ by $1$ .

$\int \frac{y^{2} - 1 y + y - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.9

Add $- 1 y$ and $y$ .

$\int \frac{y^{2} + 0 - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.10

Subtract $1$ from $0$ .

$\int \frac{y^{2} - 1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.11

Divide $y^{2} - 1$ by $y$ .

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Step 4.2.11.1

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

$y$

$0$

$y^{2}$

$0 y$

$1$

Step 4.2.11.2

Divide the highest order term in the dividend $y^{2}$ by the highest order term in divisor $y$ .

					$y$
	$y$	+	$0$		$y^{2}$	+	$0 y$	-	$1$

Step 4.2.11.3

Multiply the new quotient term by the divisor.

				$y$
$y$	+	$0$		$y^{2}$	+	$0 y$	-	$1$
			+	$y^{2}$	+	$0$

Step 4.2.11.4

The expression needs to be subtracted from the dividend, so change all the signs in $y^{2} + 0$

				$y$
$y$	+	$0$		$y^{2}$	+	$0 y$	-	$1$
			-	$y^{2}$	-	$0$

Step 4.2.11.5

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

				$y$
$y$	+	$0$		$y^{2}$	+	$0 y$	-	$1$
			-	$y^{2}$	-	$0$
						$0$

Step 4.2.11.6

Pull the next term from the original dividend down into the current dividend.

				$y$
$y$	+	$0$		$y^{2}$	+	$0 y$	-	$1$
			-	$y^{2}$	-	$0$
						$0$	-	$1$

Step 4.2.11.7

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

$\int y - \frac{1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.12

Split the single integral into multiple integrals.

$\int y d y + \int - \frac{1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.13

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

$\frac{1}{2} y^{2} + C_{1} + \int - \frac{1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.14

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

$\frac{1}{2} y^{2} + C_{1} - \int \frac{1}{y} d y = \int \frac{(x + 1) (x - 1)}{x} d x$

Step 4.2.15

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

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

Step 4.2.16

Simplify.

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Apply the distributive property.

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

Step 4.3.2

Apply the distributive property.

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

Step 4.3.3

Apply the distributive property.

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

Step 4.3.4

Reorder $x$ and $- 1$ .

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

Step 4.3.5

Raise $x$ to the power of $1$ .

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

Step 4.3.6

Raise $x$ to the power of $1$ .

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

Step 4.3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 4.3.8

Simplify the expression.

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Step 4.3.8.1

Add $1$ and $1$ .

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

Step 4.3.8.2

Multiply $x$ by $1$ .

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

Step 4.3.8.3

Multiply $- 1$ by $1$ .

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

Step 4.3.9

Add $- 1 x$ and $x$ .

$\frac{1}{2} y^{2} - \ln (| y |) + C_{3} = \int \frac{x^{2} + 0 - 1}{x} d x$

Step 4.3.10

Subtract $1$ from $0$ .

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

Step 4.3.11

Divide $x^{2} - 1$ by $x$ .

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Step 4.3.11.1

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

$x$

$0$

$x^{2}$

$0 x$

$1$

Step 4.3.11.2

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

					$x$
	$x$	+	$0$		$x^{2}$	+	$0 x$	-	$1$

Step 4.3.11.3

Multiply the new quotient term by the divisor.

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$1$
			+	$x^{2}$	+	$0$

Step 4.3.11.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + 0$

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$1$
			-	$x^{2}$	-	$0$

Step 4.3.11.5

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

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$1$
			-	$x^{2}$	-	$0$
						$0$

Step 4.3.11.6

Pull the next term from the original dividend down into the current dividend.

				$x$
$x$	+	$0$		$x^{2}$	+	$0 x$	-	$1$
			-	$x^{2}$	-	$0$
						$0$	-	$1$

Step 4.3.11.7

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

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

Step 4.3.12

Split the single integral into multiple integrals.

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

Step 4.3.13

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

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

Step 4.3.14

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

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

Step 4.3.15

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

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

Step 4.3.16

Simplify.

$\frac{1}{2} y^{2} - \ln (| y |) + C_{3} = \frac{1}{2} x^{2} - \ln (| x |) + C_{6}$

Step 4.4

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

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