Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

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

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

Step 3.3

Simplify the denominator.

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Step 3.3.1

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

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

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 $a = 1$ and $b = y$ .

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

Step 3.4

Rewrite using the commutative property of multiplication.

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

Step 3.5

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

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Step 3.5.1

Move the leading negative in $- \frac{1}{(1 - x^{2}) (1 - y^{2})}$ into the numerator.

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

Step 3.5.2

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

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

Step 3.5.3

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

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

Step 3.5.4

Cancel the common factor.

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

Step 3.5.5

Rewrite the expression.

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

Step 3.6

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

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

Step 3.7

Simplify the denominator.

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Step 3.7.1

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

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

Step 3.7.2

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

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

Step 3.8

Move the negative in front of the fraction.

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

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

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

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Step 4.2.1.1.1

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

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

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 $f (y) = 1 + y$ and $g (y) = 1 - y$ .

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

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 $1 - y$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [- y]$ .

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

Step 4.2.1.1.3.2

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

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

Step 4.2.1.1.3.3

Add $0$ and $\frac{d}{d y} [- y]$ .

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

Step 4.2.1.1.3.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

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

Step 4.2.1.1.3.5

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

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

Step 4.2.1.1.3.6

Simplify the expression.

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Step 4.2.1.1.3.6.1

Multiply $- 1$ by $1$ .

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

Step 4.2.1.1.3.6.2

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

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

Step 4.2.1.1.3.6.3

Rewrite $- 1 (1 + y)$ as $- (1 + y)$ .

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

Step 4.2.1.1.3.7

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

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

Step 4.2.1.1.3.8

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

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

Step 4.2.1.1.3.9

Add $0$ and $\frac{d}{d y} [y]$ .

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

Step 4.2.1.1.3.10

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

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

Step 4.2.1.1.3.11

Multiply $1 - y$ by $1$ .

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

Step 4.2.1.1.4

Simplify.

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Step 4.2.1.1.4.1

Apply the distributive property.

$- 1 \cdot 1 - y + 1 - y$

Step 4.2.1.1.4.2

Combine terms.

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Step 4.2.1.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - y + 1 - y$

Step 4.2.1.1.4.2.2

Add $- 1$ and $1$ .

$- y + 0 - y$

Step 4.2.1.1.4.2.3

Add $- y$ and $0$ .

$- y - y$

Step 4.2.1.1.4.2.4

Subtract $y$ from $- y$ .

$- 2 y$

Step 4.2.1.2

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

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

Step 4.2.2

Simplify.

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Step 4.2.2.1

Move the negative in front of the fraction.

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

Step 4.2.2.2

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

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

Step 4.2.2.3

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

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

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

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

Step 4.2.6

Simplify.

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

Step 4.2.7

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

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

Let $u_{2} = (1 + x) (1 - x)$ . 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 4.3.2.1

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

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Step 4.3.2.1.1

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

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

Step 4.3.2.1.2

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 $f (x) = 1 + x$ and $g (x) = 1 - x$ .

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

Step 4.3.2.1.3

Differentiate.

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Step 4.3.2.1.3.1

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

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

Step 4.3.2.1.3.2

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

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

Step 4.3.2.1.3.3

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 4.3.2.1.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 4.3.2.1.3.5

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

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

Step 4.3.2.1.3.6

Simplify the expression.

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Step 4.3.2.1.3.6.1

Multiply $- 1$ by $1$ .

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

Step 4.3.2.1.3.6.2

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

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

Step 4.3.2.1.3.6.3

Rewrite $- 1 (1 + x)$ as $- (1 + x)$ .

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

Step 4.3.2.1.3.7

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

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

Step 4.3.2.1.3.8

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

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

Step 4.3.2.1.3.9

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 4.3.2.1.3.10

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

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

Step 4.3.2.1.3.11

Multiply $1 - x$ by $1$ .

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

Step 4.3.2.1.4

Simplify.

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Step 4.3.2.1.4.1

Apply the distributive property.

$- 1 \cdot 1 - x + 1 - x$

Step 4.3.2.1.4.2

Combine terms.

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Step 4.3.2.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - x + 1 - x$

Step 4.3.2.1.4.2.2

Add $- 1$ and $1$ .

$- x + 0 - x$

Step 4.3.2.1.4.2.3

Add $- x$ and $0$ .

$- x - x$

Step 4.3.2.1.4.2.4

Subtract $x$ from $- x$ .

$- 2 x$

Step 4.3.2.2

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

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

Step 4.3.3

Simplify.

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Step 4.3.3.1

Move the negative in front of the fraction.

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

Step 4.3.3.2

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

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

Step 4.3.3.3

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

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

Step 4.3.4

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

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

Step 4.3.5

Simplify.

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Step 4.3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.5.2

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

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

Step 4.3.6

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

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

Step 4.3.7

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

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

Step 4.3.8

Simplify.

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

Step 4.3.9

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $- 2$ .

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

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

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Step 5.2.1.1.1

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

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Step 5.2.1.1.1.1

Apply the distributive property.

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

Step 5.2.1.1.1.2

Apply the distributive property.

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

Step 5.2.1.1.1.3

Apply the distributive property.

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

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 $1$ by $1$ .

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

Step 5.2.1.1.2.1.2

Multiply $- y$ by $1$ .

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

Step 5.2.1.1.2.1.3

Multiply $y$ by $1$ .

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

Step 5.2.1.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 5.2.1.1.2.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 5.2.1.1.2.1.5.1

Move $y$ .

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

Step 5.2.1.1.2.1.5.2

Multiply $y$ by $y$ .

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

Step 5.2.1.1.2.2

Add $- y$ and $y$ .

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

Step 5.2.1.1.2.3

Add $1$ and $0$ .

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

Step 5.2.1.1.3

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

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

Step 5.2.1.1.4

Cancel the common factor of $2$ .

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Step 5.2.1.1.4.1

Move the leading negative in $- \frac{\ln (| 1 - y^{2} |)}{2}$ into the numerator.

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

Step 5.2.1.1.4.2

Factor $2$ out of $- 2$ .

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

Step 5.2.1.1.4.3

Cancel the common factor.

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

Step 5.2.1.1.4.4

Rewrite the expression.

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

Step 5.2.1.1.5

Multiply.

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Step 5.2.1.1.5.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.1.1.5.2

Multiply $\ln (| 1 - y^{2} |)$ by $1$ .

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

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

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Step 5.2.2.1.1.1.1

Apply the distributive property.

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

Step 5.2.2.1.1.1.2

Apply the distributive property.

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

Step 5.2.2.1.1.1.3

Apply the distributive property.

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

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 $1$ by $1$ .

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

Step 5.2.2.1.1.2.1.2

Multiply $- x$ by $1$ .

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

Step 5.2.2.1.1.2.1.3

Multiply $x$ by $1$ .

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

Step 5.2.2.1.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 5.2.2.1.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 5.2.2.1.1.2.1.5.1

Move $x$ .

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

Step 5.2.2.1.1.2.1.5.2

Multiply $x$ by $x$ .

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

Step 5.2.2.1.1.2.2

Add $- x$ and $x$ .

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

Step 5.2.2.1.1.2.3

Add $1$ and $0$ .

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

Step 5.2.2.1.1.3

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

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

Step 5.2.2.1.2

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

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

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

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

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

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

Factor $2$ out of $- 2$ .

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

Step 5.2.2.1.3.3.2

Cancel the common factor.

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

Step 5.2.2.1.3.3.3

Rewrite the expression.

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

Step 5.2.2.1.4

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

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

Step 5.2.2.1.5

Simplify by multiplying through.

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Step 5.2.2.1.5.1

Apply the distributive property.

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

Step 5.2.2.1.5.2

Multiply $2$ by $- 1$ .

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

Step 5.3

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

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

Step 5.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 5.5

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 5.6

Expand $(1 - y^{2}) (1 - x^{2})$ using the FOIL Method.

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Step 5.6.1

Apply the distributive property.

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

Step 5.6.2

Apply the distributive property.

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

Step 5.6.3

Apply the distributive property.

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

Step 5.7

Simplify each term.

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Step 5.7.1

Multiply $1$ by $1$ .

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

Step 5.7.2

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

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

Step 5.7.3

Multiply $- 1$ by $1$ .

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

Step 5.7.4

Rewrite using the commutative property of multiplication.

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

Step 5.7.5

Multiply $- 1$ by $- 1$ .

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

Step 5.7.6

Multiply $y^{2}$ by $1$ .

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

Step 5.8

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

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

Step 5.9

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

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

Step 5.10

Solve for $y$ .

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Step 5.10.1

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

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

Step 5.10.2

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

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

Step 5.10.3

Move all terms not containing $y$ to the right side of the equation.

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Step 5.10.3.1

Subtract $1$ from both sides of the equation.

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

Step 5.10.3.2

Add $x^{2}$ to both sides of the equation.

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

Step 5.10.4

Factor $y^{2}$ out of $- y^{2} + y^{2} x^{2}$ .

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Step 5.10.4.1

Factor $y^{2}$ out of $- y^{2}$ .

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

Step 5.10.4.2

Factor $y^{2}$ out of $y^{2} x^{2}$ .

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

Step 5.10.4.3

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

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

Step 5.10.5

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

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

Step 5.10.6

Reorder $- 1^{2}$ and $x^{2}$ .

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

Step 5.10.7

Factor.

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Step 5.10.7.1

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 $b = 1$ .

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

Step 5.10.7.2

Remove unnecessary parentheses.

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

Step 5.10.8

Divide each term in $y^{2} (x + 1) (x - 1) = \pm e^{- 2 C} - 1 + x^{2}$ by $(x + 1) (x - 1)$ and simplify.

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Step 5.10.8.1

Divide each term in $y^{2} (x + 1) (x - 1) = \pm e^{- 2 C} - 1 + x^{2}$ by $(x + 1) (x - 1)$ .

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

Step 5.10.8.2

Simplify the left side.

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Step 5.10.8.2.1

Cancel the common factor of $x + 1$ .

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Step 5.10.8.2.1.1

Cancel the common factor.

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

Step 5.10.8.2.1.2

Rewrite the expression.

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

Step 5.10.8.2.2

Cancel the common factor of $x - 1$ .

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Step 5.10.8.2.2.1

Cancel the common factor.

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

Step 5.10.8.2.2.2

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

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

Step 5.10.8.3

Simplify the right side.

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Step 5.10.8.3.1

Move the negative in front of the fraction.

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

Step 5.10.9

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

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

Step 5.10.10

Simplify $\pm \sqrt{\frac{\pm e^{- 2 C}}{(x + 1) (x - 1)} - \frac{1}{(x + 1) (x - 1)} + \frac{x^{2}}{(x + 1) (x - 1)}}$ .

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Step 5.10.10.1

Combine the numerators over the common denominator.

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

Step 5.10.10.2

Combine the numerators over the common denominator.

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

Step 5.10.10.3

Rewrite $\sqrt{\frac{\pm e^{- 2 C} - 1 + x^{2}}{(x + 1) (x - 1)}}$ as $\frac{\sqrt{\pm e^{- 2 C} - 1 + x^{2}}}{\sqrt{(x + 1) (x - 1)}}$ .

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

Step 5.10.10.4

Multiply $\frac{\sqrt{\pm e^{- 2 C} - 1 + x^{2}}}{\sqrt{(x + 1) (x - 1)}}$ by $\frac{\sqrt{(x + 1) (x - 1)}}{\sqrt{(x + 1) (x - 1)}}$ .

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

Step 5.10.10.5

Combine and simplify the denominator.

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Step 5.10.10.5.1

Multiply $\frac{\sqrt{\pm e^{- 2 C} - 1 + x^{2}}}{\sqrt{(x + 1) (x - 1)}}$ by $\frac{\sqrt{(x + 1) (x - 1)}}{\sqrt{(x + 1) (x - 1)}}$ .

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

Step 5.10.10.5.2

Raise $\sqrt{(x + 1) (x - 1)}$ to the power of $1$ .

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

Step 5.10.10.5.3

Raise $\sqrt{(x + 1) (x - 1)}$ to the power of $1$ .

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

Step 5.10.10.5.4

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

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

Step 5.10.10.5.5

Add $1$ and $1$ .

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

Step 5.10.10.5.6

Rewrite ${\sqrt{(x + 1) (x - 1)}}^{2}$ as $(x + 1) (x - 1)$ .

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Step 5.10.10.5.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(x + 1) (x - 1)}$ as ${((x + 1) (x - 1))}^{\frac{1}{2}}$ .

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

Step 5.10.10.5.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 5.10.10.5.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 5.10.10.5.6.4

Cancel the common factor of $2$ .

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Step 5.10.10.5.6.4.1

Cancel the common factor.

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

Step 5.10.10.5.6.4.2

Rewrite the expression.

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

Step 5.10.10.5.6.5

Simplify.

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

Step 5.10.10.6

Combine using the product rule for radicals.

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

Step 6

Simplify the constant of integration.

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