Calculus Examples

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

Step 1

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

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

Step 2

Multiply both sides by $\frac{1}{(x^{2} - 1) (y^{2} - 1)}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x^{2} - 1$ .

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Step 3.1.1

Factor $x^{2} - 1$ out of $y (x^{2} - 1)$ .

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Combine $\frac{1}{y^{2} - 1}$ and $y$ .

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

Step 3.3

Simplify the denominator.

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Step 3.3.1

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

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

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 $b = 1$ .

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

Step 3.4

Cancel the common factor of $y^{2} - 1$ .

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Step 3.4.1

Factor $y^{2} - 1$ out of $(x^{2} - 1) (y^{2} - 1)$ .

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

Step 3.4.2

Factor $y^{2} - 1$ out of $x (y^{2} - 1)$ .

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

Step 3.4.3

Cancel the common factor.

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

Step 3.4.4

Rewrite the expression.

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

Step 3.5

Combine $\frac{1}{x^{2} - 1}$ and $x$ .

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

Step 3.6

Simplify the denominator.

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Step 3.6.1

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

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

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 $b = 1$ .

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u_{1} = (y + 1) (y - 1)$ . Then $d u_{1} = 2 y d y$ , so $\frac{1}{2} d u_{1} = y d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.2.1.1

Let $u_{1} = (y + 1) (y - 1)$ . Find $\frac{d u_{1}}{d y}$ .

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Step 4.2.1.1.1

Differentiate $(y + 1) (y - 1)$ .

$\frac{d}{d y} [(y + 1) (y - 1)]$

Step 4.2.1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y + 1$ and $g (y) = y - 1$ .

$(y + 1) \frac{d}{d y} [y - 1] + (y - 1) \frac{d}{d y} [y + 1]$

Step 4.2.1.1.3

Differentiate.

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Step 4.2.1.1.3.1

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

$(y + 1) (\frac{d}{d y} [y] + \frac{d}{d y} [- 1]) + (y - 1) \frac{d}{d y} [y + 1]$

Step 4.2.1.1.3.2

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

$(y + 1) (1 + \frac{d}{d y} [- 1]) + (y - 1) \frac{d}{d y} [y + 1]$

Step 4.2.1.1.3.3

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

$(y + 1) (1 + 0) + (y - 1) \frac{d}{d y} [y + 1]$

Step 4.2.1.1.3.4

Simplify the expression.

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Step 4.2.1.1.3.4.1

Add $1$ and $0$ .

$(y + 1) \cdot 1 + (y - 1) \frac{d}{d y} [y + 1]$

Step 4.2.1.1.3.4.2

Multiply $y + 1$ by $1$ .

$y + 1 + (y - 1) \frac{d}{d y} [y + 1]$

Step 4.2.1.1.3.5

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

$y + 1 + (y - 1) (\frac{d}{d y} [y] + \frac{d}{d y} [1])$

Step 4.2.1.1.3.6

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

$y + 1 + (y - 1) (1 + \frac{d}{d y} [1])$

Step 4.2.1.1.3.7

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

$y + 1 + (y - 1) (1 + 0)$

Step 4.2.1.1.3.8

Simplify by adding terms.

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Step 4.2.1.1.3.8.1

Add $1$ and $0$ .

$y + 1 + (y - 1) \cdot 1$

Step 4.2.1.1.3.8.2

Multiply $y - 1$ by $1$ .

$y + 1 + y - 1$

Step 4.2.1.1.3.8.3

Add $y$ and $y$ .

$2 y + 1 - 1$

Step 4.2.1.1.3.8.4

Simplify by subtracting numbers.

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Step 4.2.1.1.3.8.4.1

Subtract $1$ from $1$ .

$2 y + 0$

Step 4.2.1.1.3.8.4.2

Add $2 y$ and $0$ .

$2 y$

Step 4.2.1.2

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

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

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

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

Step 4.2.2.2

Move $2$ to the left of $u_{1}$ .

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

Replace all occurrences of $u_{1}$ with $(y + 1) (y - 1)$ .

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Let $u_{2} = (x + 1) (x - 1)$ . Then $d u_{2} = 2 x d x$ , so $\frac{1}{2} d u_{2} = x d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.1.1

Let $u_{2} = (x + 1) (x - 1)$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.1.1.1

Differentiate $(x + 1) (x - 1)$ .

$\frac{d}{d x} [(x + 1) (x - 1)]$

Step 4.3.1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = x - 1$ .

$(x + 1) \frac{d}{d x} [x - 1] + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.1.1.3

Differentiate.

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Step 4.3.1.1.3.1

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

$(x + 1) (\frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.1.1.3.2

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

$(x + 1) (1 + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.1.1.3.3

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

$(x + 1) (1 + 0) + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.1.1.3.4

Simplify the expression.

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Step 4.3.1.1.3.4.1

Add $1$ and $0$ .

$(x + 1) \cdot 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.1.1.3.4.2

Multiply $x + 1$ by $1$ .

$x + 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 4.3.1.1.3.5

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

$x + 1 + (x - 1) (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 4.3.1.1.3.6

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

$x + 1 + (x - 1) (1 + \frac{d}{d x} [1])$

Step 4.3.1.1.3.7

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

$x + 1 + (x - 1) (1 + 0)$

Step 4.3.1.1.3.8

Simplify by adding terms.

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Step 4.3.1.1.3.8.1

Add $1$ and $0$ .

$x + 1 + (x - 1) \cdot 1$

Step 4.3.1.1.3.8.2

Multiply $x - 1$ by $1$ .

$x + 1 + x - 1$

Step 4.3.1.1.3.8.3

Add $x$ and $x$ .

$2 x + 1 - 1$

Step 4.3.1.1.3.8.4

Simplify by subtracting numbers.

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Step 4.3.1.1.3.8.4.1

Subtract $1$ from $1$ .

$2 x + 0$

Step 4.3.1.1.3.8.4.2

Add $2 x$ and $0$ .

$2 x$

Step 4.3.1.2

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

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

Step 4.3.2

Simplify.

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Step 4.3.2.1

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

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

Step 4.3.2.2

Move $2$ to the left of $u_{2}$ .

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

Step 4.3.3

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

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

Step 4.3.4

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

$\frac{1}{2} \ln (| (y + 1) (y - 1) |) + C_{1} = \frac{1}{2} (\ln (| u_{2} |) + C_{2})$

Step 4.3.5

Simplify.

$\frac{1}{2} \ln (| (y + 1) (y - 1) |) + C_{1} = \frac{1}{2} \ln (| u_{2} |) + C_{2}$

Step 4.3.6

Replace all occurrences of $u_{2}$ with $(x + 1) (x - 1)$ .

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Simplify $2 (\frac{1}{2} \ln (| (y + 1) (y - 1) |))$ .

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Step 5.2.1.1.1

Expand $(y + 1) (y - 1)$ using the FOIL Method.

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Step 5.2.1.1.1.1

Apply the distributive property.

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

Step 5.2.1.1.1.2

Apply the distributive property.

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

Step 5.2.1.1.1.3

Apply the distributive property.

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

Step 5.2.1.1.2

Simplify and combine like terms.

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Step 5.2.1.1.2.1

Simplify each term.

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Step 5.2.1.1.2.1.1

Multiply $y$ by $y$ .

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

Step 5.2.1.1.2.1.2

Move $- 1$ to the left of $y$ .

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

Step 5.2.1.1.2.1.3

Rewrite $- 1 y$ as $- y$ .

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

Step 5.2.1.1.2.1.4

Multiply $y$ by $1$ .

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

Step 5.2.1.1.2.1.5

Multiply $- 1$ by $1$ .

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

Step 5.2.1.1.2.2

Add $- y$ and $y$ .

$2 (\frac{1}{2} \ln (| y^{2} + 0 - 1 |)) = 2 (\frac{1}{2} \ln (| (x + 1) (x - 1) |) + C)$

Step 5.2.1.1.2.3

Add $y^{2}$ and $0$ .

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

Step 5.2.1.1.3

Combine $\frac{1}{2}$ and $\ln (| y^{2} - 1 |)$ .

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

Step 5.2.1.1.4

Cancel the common factor of $2$ .

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Step 5.2.1.1.4.1

Cancel the common factor.

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

Step 5.2.1.1.4.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $2 (\frac{1}{2} \ln (| (x + 1) (x - 1) |) + C)$ .

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Step 5.2.2.1.1

Simplify each term.

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Step 5.2.2.1.1.1

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 5.2.2.1.1.1.1

Apply the distributive property.

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

Step 5.2.2.1.1.1.2

Apply the distributive property.

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

Step 5.2.2.1.1.1.3

Apply the distributive property.

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

Step 5.2.2.1.1.2

Simplify and combine like terms.

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Step 5.2.2.1.1.2.1

Simplify each term.

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Step 5.2.2.1.1.2.1.1

Multiply $x$ by $x$ .

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

Step 5.2.2.1.1.2.1.2

Move $- 1$ to the left of $x$ .

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

Step 5.2.2.1.1.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 5.2.2.1.1.2.1.4

Multiply $x$ by $1$ .

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

Step 5.2.2.1.1.2.1.5

Multiply $- 1$ by $1$ .

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

Step 5.2.2.1.1.2.2

Add $- x$ and $x$ .

$\ln (| y^{2} - 1 |) = 2 (\frac{1}{2} \ln (| x^{2} + 0 - 1 |) + C)$

Step 5.2.2.1.1.2.3

Add $x^{2}$ and $0$ .

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

Step 5.2.2.1.1.3

Combine $\frac{1}{2}$ and $\ln (| x^{2} - 1 |)$ .

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

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

Combine $C$ and $\frac{2}{2}$ .

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

Step 5.2.2.1.3.3

Cancel the common factor of $2$ .

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Step 5.2.2.1.3.3.1

Cancel the common factor.

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

Step 5.2.2.1.3.3.2

Rewrite the expression.

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

Step 5.2.2.1.4

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

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

Step 5.3

Move all the terms containing a logarithm to the left side of the equation.

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

Step 5.4

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 5.5

Simplify the numerator.

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Step 5.5.1

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

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

Step 5.5.2

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 $b = 1$ .

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

Step 5.5.3

Expand $(y + 1) (y - 1)$ using the FOIL Method.

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Step 5.5.3.1

Apply the distributive property.

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

Step 5.5.3.2

Apply the distributive property.

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

Step 5.5.3.3

Apply the distributive property.

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

Step 5.5.4

Simplify and combine like terms.

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Step 5.5.4.1

Simplify each term.

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Step 5.5.4.1.1

Multiply $y$ by $y$ .

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

Step 5.5.4.1.2

Move $- 1$ to the left of $y$ .

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

Step 5.5.4.1.3

Rewrite $- 1 y$ as $- y$ .

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

Step 5.5.4.1.4

Multiply $y$ by $1$ .

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

Step 5.5.4.1.5

Multiply $- 1$ by $1$ .

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

Step 5.5.4.2

Add $- y$ and $y$ .

$\ln (\frac{| y^{2} + 0 - 1 |}{| x^{2} - 1 |}) = 2 C$

Step 5.5.4.3

Add $y^{2}$ and $0$ .

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

Step 5.5.5

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

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

Step 5.5.6

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 $b = 1$ .

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

Step 5.6

Simplify the denominator.

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Step 5.6.1

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

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

Step 5.6.2

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 $b = 1$ .

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

Step 5.6.3

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 5.6.3.1

Apply the distributive property.

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

Step 5.6.3.2

Apply the distributive property.

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

Step 5.6.3.3

Apply the distributive property.

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

Step 5.6.4

Simplify and combine like terms.

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Step 5.6.4.1

Simplify each term.

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Step 5.6.4.1.1

Multiply $x$ by $x$ .

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

Step 5.6.4.1.2

Move $- 1$ to the left of $x$ .

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

Step 5.6.4.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 5.6.4.1.4

Multiply $x$ by $1$ .

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

Step 5.6.4.1.5

Multiply $- 1$ by $1$ .

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

Step 5.6.4.2

Add $- x$ and $x$ .

$\ln (\frac{| (y + 1) (y - 1) |}{| x^{2} + 0 - 1 |}) = 2 C$

Step 5.6.4.3

Add $x^{2}$ and $0$ .

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

Step 5.6.5

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

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

Step 5.6.6

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 $b = 1$ .

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

Step 5.7

To solve for $y$ , rewrite the equation using properties of logarithms.

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

Step 5.8

Rewrite $\ln (\frac{| (y + 1) (y - 1) |}{| (x + 1) (x - 1) |}) = 2 C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Step 5.9

Solve for $y$ .

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Step 5.9.1

Rewrite the equation as $\frac{| (y + 1) (y - 1) |}{| (x + 1) (x - 1) |} = e^{2 C}$ .

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

Step 5.9.2

Multiply both sides by $| (x + 1) (x - 1) |$ .

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

Step 5.9.3

Simplify.

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Step 5.9.3.1

Simplify the left side.

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Step 5.9.3.1.1

Simplify $\frac{| (y + 1) (y - 1) |}{| (x + 1) (x - 1) |} | (x + 1) (x - 1) |$ .

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Step 5.9.3.1.1.1

Cancel the common factor of $| (x + 1) (x - 1) |$ .

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Step 5.9.3.1.1.1.1

Cancel the common factor.

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

Step 5.9.3.1.1.1.2

Rewrite the expression.

$| (y + 1) (y - 1) | = e^{2 C} | (x + 1) (x - 1) |$

Step 5.9.3.1.1.2

Expand $(y + 1) (y - 1)$ using the FOIL Method.

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Step 5.9.3.1.1.2.1

Apply the distributive property.

$| y (y - 1) + 1 (y - 1) | = e^{2 C} | (x + 1) (x - 1) |$

Step 5.9.3.1.1.2.2

Apply the distributive property.

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

Step 5.9.3.1.1.2.3

Apply the distributive property.

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

Step 5.9.3.1.1.3

Simplify and combine like terms.

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Step 5.9.3.1.1.3.1

Simplify each term.

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Step 5.9.3.1.1.3.1.1

Multiply $y$ by $y$ .

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

Step 5.9.3.1.1.3.1.2

Move $- 1$ to the left of $y$ .

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

Step 5.9.3.1.1.3.1.3

Rewrite $- 1 y$ as $- y$ .

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

Step 5.9.3.1.1.3.1.4

Multiply $y$ by $1$ .

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

Step 5.9.3.1.1.3.1.5

Multiply $- 1$ by $1$ .

$| y^{2} - y + y - 1 | = e^{2 C} | (x + 1) (x - 1) |$

Step 5.9.3.1.1.3.2

Add $- y$ and $y$ .

$| y^{2} + 0 - 1 | = e^{2 C} | (x + 1) (x - 1) |$

Step 5.9.3.1.1.3.3

Add $y^{2}$ and $0$ .

$| y^{2} - 1 | = e^{2 C} | (x + 1) (x - 1) |$

Step 5.9.3.2

Simplify the right side.

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Step 5.9.3.2.1

Simplify $e^{2 C} | (x + 1) (x - 1) |$ .

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Step 5.9.3.2.1.1

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 5.9.3.2.1.1.1

Apply the distributive property.

$| y^{2} - 1 | = e^{2 C} | x (x - 1) + 1 (x - 1) |$

Step 5.9.3.2.1.1.2

Apply the distributive property.

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

Step 5.9.3.2.1.1.3

Apply the distributive property.

$| y^{2} - 1 | = e^{2 C} | x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1 |$

Step 5.9.3.2.1.2

Simplify and combine like terms.

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Step 5.9.3.2.1.2.1

Simplify each term.

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Step 5.9.3.2.1.2.1.1

Multiply $x$ by $x$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1 |$

Step 5.9.3.2.1.2.1.2

Move $- 1$ to the left of $x$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1 |$

Step 5.9.3.2.1.2.1.3

Rewrite $- 1 x$ as $- x$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} - x + 1 x + 1 \cdot - 1 |$

Step 5.9.3.2.1.2.1.4

Multiply $x$ by $1$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} - x + x + 1 \cdot - 1 |$

Step 5.9.3.2.1.2.1.5

Multiply $- 1$ by $1$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} - x + x - 1 |$

Step 5.9.3.2.1.2.2

Add $- x$ and $x$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} + 0 - 1 |$

Step 5.9.3.2.1.2.3

Add $x^{2}$ and $0$ .

$| y^{2} - 1 | = e^{2 C} | x^{2} - 1 |$

Step 5.9.4

Solve for $y$ .

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Step 5.9.4.1

Reorder factors in $e^{2 C} | x^{2} - 1 |$ .

$| y^{2} - 1 | = | x^{2} - 1 | e^{2 C}$

Step 5.9.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y^{2} - 1 = \pm | x^{2} - 1 | e^{2 C}$

Step 5.9.4.3

Reorder factors in $\pm | x^{2} - 1 | e^{2 C}$ .

$y^{2} - 1 = \pm e^{2 C} | x^{2} - 1 |$

Step 5.9.4.4

Add $1$ to both sides of the equation.

$y^{2} = \pm e^{2 C} | x^{2} - 1 | + 1$

Step 5.9.4.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\pm e^{2 C} | x^{2} - 1 | + 1}$

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

$y = \pm \sqrt{\pm C | x^{2} - 1 | + 1}$

Step 6.2

Combine constants with the plus or minus.

$y = \pm \sqrt{C | x^{2} - 1 | + 1}$