Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Simplify each term.

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Step 1.1.1.1

Apply the distributive property.

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

Step 1.1.1.2

Multiply $\frac{d y}{d x}$ by $1$ .

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

Step 1.1.2

Subtract $x y$ from both sides of the equation.

$\frac{d y}{d x} - x^{2} \frac{d y}{d x} = 3 x - x y$

Step 1.1.3

Factor $\frac{d y}{d x}$ out of $\frac{d y}{d x} - x^{2} \frac{d y}{d x}$ .

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Step 1.1.3.1

Factor $\frac{d y}{d x}$ out of ${(\frac{d y}{d x})}^{1}$ .

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

Step 1.1.3.2

Factor $\frac{d y}{d x}$ out of $- x^{2} \frac{d y}{d x}$ .

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

Step 1.1.3.3

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

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

Step 1.1.4

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

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

Step 1.1.5

Factor.

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Step 1.1.5.1

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 $b = x$ .

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

Step 1.1.5.2

Remove unnecessary parentheses.

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

Step 1.1.6

Divide each term in $\frac{d y}{d x} (1 + x) (1 - x) = 3 x - x y$ by $(1 + x) (1 - x)$ and simplify.

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Step 1.1.6.1

Divide each term in $\frac{d y}{d x} (1 + x) (1 - x) = 3 x - x y$ by $(1 + x) (1 - x)$ .

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

Step 1.1.6.2

Simplify the left side.

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Step 1.1.6.2.1

Cancel the common factor of $1 + x$ .

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Step 1.1.6.2.1.1

Cancel the common factor.

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

Step 1.1.6.2.1.2

Rewrite the expression.

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

Step 1.1.6.2.2

Cancel the common factor of $1 - x$ .

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Step 1.1.6.2.2.1

Cancel the common factor.

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

Step 1.1.6.2.2.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.1.6.3

Simplify the right side.

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Step 1.1.6.3.1

Move the negative in front of the fraction.

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

Step 1.2

Factor.

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Step 1.2.1

Combine the numerators over the common denominator.

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

Step 1.2.2

Simplify the numerator.

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Step 1.2.2.1

Factor $x$ out of $3 x - x y$ .

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Step 1.2.2.1.1

Factor $x$ out of $3 x$ .

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

Step 1.2.2.1.2

Factor $x$ out of $- x y$ .

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

Step 1.2.2.1.3

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

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

Step 1.2.2.1.4

Multiply $x$ by $3 - 1 y$ .

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

Step 1.2.2.2

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

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

Step 1.3

Regroup factors.

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

Step 1.4

Multiply both sides by $\frac{1}{3 - y}$ .

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

Step 1.5

Cancel the common factor of $3 - y$ .

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Step 1.5.1

Factor $3 - y$ out of $\frac{x}{(1 + x) (1 - x)} (3 - y)$ .

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

Step 1.5.2

Cancel the common factor.

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

Step 1.5.3

Rewrite the expression.

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

Step 1.6

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = 3 - y$ . Then $d u_{1} = - d y$ , so $- d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

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

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

Step 2.2.2

Split the fraction into multiple fractions.

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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 2.3.1.1

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

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Step 2.3.1.1.1

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

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

Differentiate.

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Step 2.3.1.1.3.1

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 2.3.1.1.3.2

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 2.3.1.1.3.3

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

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

Step 2.3.1.1.3.4

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 2.3.1.1.3.5

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 2.3.1.1.3.6

Simplify the expression.

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Step 2.3.1.1.3.6.1

Multiply $- 1$ by $1$ .

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

Step 2.3.1.1.3.6.2

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

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

Step 2.3.1.1.3.6.3

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

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

Step 2.3.1.1.3.7

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 2.3.1.1.3.8

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 2.3.1.1.3.9

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

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

Step 2.3.1.1.3.10

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 2.3.1.1.3.11

Multiply $1 - x$ by $1$ .

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

Step 2.3.1.1.4

Simplify.

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Step 2.3.1.1.4.1

Apply the distributive property.

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

Step 2.3.1.1.4.2

Combine terms.

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Step 2.3.1.1.4.2.1

Multiply $- 1$ by $1$ .

$- 1 - x + 1 - x$

Step 2.3.1.1.4.2.2

Add $- 1$ and $1$ .

$- x + 0 - x$

Step 2.3.1.1.4.2.3

Add $- x$ and $0$ .

$- x - x$

Step 2.3.1.1.4.2.4

Subtract $x$ from $- x$ .

$- 2 x$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

Move the negative in front of the fraction.

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Simplify.

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

Step 2.3.7

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

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

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

Step 3.2

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

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

Step 3.3

Simplify the numerator.

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Step 3.3.1

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

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Step 3.3.1.1

Apply the distributive property.

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

Step 3.3.1.2

Apply the distributive property.

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

Step 3.3.1.3

Apply the distributive property.

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

Step 3.3.2

Simplify and combine like terms.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $1$ by $1$ .

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

Step 3.3.2.1.2

Multiply $- x$ by $1$ .

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

Step 3.3.2.1.3

Multiply $x$ by $1$ .

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

Step 3.3.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 3.3.2.1.5

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

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Step 3.3.2.1.5.1

Move $x$ .

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

Step 3.3.2.1.5.2

Multiply $x$ by $x$ .

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

Step 3.3.2.2

Add $- x$ and $x$ .

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

Step 3.3.2.3

Add $1$ and $0$ .

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

Step 3.4

To write $- \ln (| 3 - y |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.5

Combine $- \ln (| 3 - y |)$ and $\frac{2}{2}$ .

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

Step 3.6

Combine the numerators over the common denominator.

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

Step 3.7

Multiply $2$ by $- 1$ .

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

Step 3.8

Factor $- 1$ out of $- 2 \ln (| 3 - y |)$ .

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

Step 3.9

Factor $- 1$ out of $\ln (| 1 - x^{2} |)$ .

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

Step 3.10

Factor $- 1$ out of $- (2 \ln (| 3 - y |)) - 1 (- \ln (| 1 - x^{2} |))$ .

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

Step 3.11

Simplify the expression.

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Step 3.11.1

Rewrite $- (2 \ln (| 3 - y |) - \ln (| 1 - x^{2} |))$ as $- 1 (2 \ln (| 3 - y |) - \ln (| 1 - x^{2} |))$ .

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

Step 3.11.2

Move the negative in front of the fraction.

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

Step 3.12

Simplify the left side.

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Step 3.12.1

Simplify $- \frac{2 \ln (| 3 - y |) - \ln (| 1 - x^{2} |)}{2}$ .

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Step 3.12.1.1

Simplify the numerator.

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Step 3.12.1.1.1

Simplify $2 \ln (| 3 - y |)$ by moving $2$ inside the logarithm.

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

Step 3.12.1.1.2

Remove the absolute value in ${| 3 - y |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.12.1.1.3

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

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

Step 3.12.1.1.4

Simplify the denominator.

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Step 3.12.1.1.4.1

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

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

Step 3.12.1.1.4.2

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 $b = x$ .

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

Step 3.12.1.1.4.3

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

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Step 3.12.1.1.4.3.1

Apply the distributive property.

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

Step 3.12.1.1.4.3.2

Apply the distributive property.

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

Step 3.12.1.1.4.3.3

Apply the distributive property.

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

Step 3.12.1.1.4.4

Simplify and combine like terms.

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Step 3.12.1.1.4.4.1

Simplify each term.

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Step 3.12.1.1.4.4.1.1

Multiply $1$ by $1$ .

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

Step 3.12.1.1.4.4.1.2

Multiply $- x$ by $1$ .

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

Step 3.12.1.1.4.4.1.3

Multiply $x$ by $1$ .

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

Step 3.12.1.1.4.4.1.4

Rewrite using the commutative property of multiplication.

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

Step 3.12.1.1.4.4.1.5

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

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Step 3.12.1.1.4.4.1.5.1

Move $x$ .

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

Step 3.12.1.1.4.4.1.5.2

Multiply $x$ by $x$ .

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

Step 3.12.1.1.4.4.2

Add $- x$ and $x$ .

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

Step 3.12.1.1.4.4.3

Add $1$ and $0$ .

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

Step 3.12.1.1.4.5

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

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

Step 3.12.1.1.4.6

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 $b = x$ .

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

Step 3.12.1.2

Rewrite $\frac{\ln (\frac{{(3 - y)}^{2}}{| (1 + x) (1 - x) |})}{2}$ as $\frac{1}{2} \ln (\frac{{(3 - y)}^{2}}{| (1 + x) (1 - x) |})$ .

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

Step 3.12.1.3

Simplify $\frac{1}{2} \ln (\frac{{(3 - y)}^{2}}{| (1 + x) (1 - x) |})$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 3.12.1.4

Apply the product rule to $\frac{{(3 - y)}^{2}}{| (1 + x) (1 - x) |}$ .

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

Step 3.12.1.5

Simplify the numerator.

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Step 3.12.1.5.1

Multiply the exponents in ${({(3 - y)}^{2})}^{\frac{1}{2}}$ .

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Step 3.12.1.5.1.1

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

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

Step 3.12.1.5.1.2

Cancel the common factor of $2$ .

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Step 3.12.1.5.1.2.1

Cancel the common factor.

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

Step 3.12.1.5.1.2.2

Rewrite the expression.

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

Step 3.12.1.5.2

Simplify.

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

Step 3.12.1.6

Simplify the denominator.

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Step 3.12.1.6.1

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

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Step 3.12.1.6.1.1

Apply the distributive property.

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

Step 3.12.1.6.1.2

Apply the distributive property.

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

Step 3.12.1.6.1.3

Apply the distributive property.

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

Step 3.12.1.6.2

Simplify and combine like terms.

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Step 3.12.1.6.2.1

Simplify each term.

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Step 3.12.1.6.2.1.1

Multiply $1$ by $1$ .

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

Step 3.12.1.6.2.1.2

Multiply $- x$ by $1$ .

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

Step 3.12.1.6.2.1.3

Multiply $x$ by $1$ .

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

Step 3.12.1.6.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 3.12.1.6.2.1.5

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

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Step 3.12.1.6.2.1.5.1

Move $x$ .

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

Step 3.12.1.6.2.1.5.2

Multiply $x$ by $x$ .

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

Step 3.12.1.6.2.2

Add $- x$ and $x$ .

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

Step 3.12.1.6.2.3

Add $1$ and $0$ .

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

Step 3.13

Divide each term in $- \ln (\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}}) = C$ by $- 1$ and simplify.

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Step 3.13.1

Divide each term in $- \ln (\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}}) = C$ by $- 1$ .

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

Step 3.13.2

Simplify the left side.

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Step 3.13.2.1

Dividing two negative values results in a positive value.

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

Step 3.13.2.2

Divide $\ln (\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}})$ by $1$ .

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

Step 3.13.3

Simplify the right side.

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Step 3.13.3.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

$\ln (\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}}) = - 1 \cdot C$

Step 3.13.3.2

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 3.14

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

$e^{\ln (\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}})} = e^{- C}$

Step 3.15

Rewrite $\ln (\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}}) = - C$ 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 $b^{y} = x$ .

$e^{- C} = \frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}}$

Step 3.16

Solve for $y$ .

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Step 3.16.1

Rewrite the equation as $\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}} = e^{- C}$ .

$\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}} = e^{- C}$

Step 3.16.2

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

$\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}} {| 1 - x^{2} |}^{\frac{1}{2}} = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}$

Step 3.16.3

Simplify the left side.

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Step 3.16.3.1

Simplify $\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}} {| 1 - x^{2} |}^{\frac{1}{2}}$ .

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Step 3.16.3.1.1

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

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Step 3.16.3.1.1.1

Cancel the common factor.

$\frac{3 - y}{{| 1 - x^{2} |}^{\frac{1}{2}}} {| 1 - x^{2} |}^{\frac{1}{2}} = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}$

Step 3.16.3.1.1.2

Rewrite the expression.

$3 - y = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}$

Step 3.16.3.1.2

Reorder $3$ and $- y$ .

$- y + 3 = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}$

Step 3.16.4

Solve for $y$ .

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Step 3.16.4.1

Subtract $3$ from both sides of the equation.

$- y = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}} - 3$

Step 3.16.4.2

Divide each term in $- y = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}} - 3$ by $- 1$ and simplify.

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Step 3.16.4.2.1

Divide each term in $- y = e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}} - 3$ by $- 1$ .

$\frac{- y}{- 1} = \frac{e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}}{- 1} + \frac{- 3}{- 1}$

Step 3.16.4.2.2

Simplify the left side.

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Step 3.16.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}}{- 1} + \frac{- 3}{- 1}$

Step 3.16.4.2.2.2

Divide $y$ by $1$ .

$y = \frac{e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}}{- 1} + \frac{- 3}{- 1}$

Step 3.16.4.2.3

Simplify the right side.

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Step 3.16.4.2.3.1

Simplify each term.

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Step 3.16.4.2.3.1.1

Move the negative one from the denominator of $\frac{e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}}{- 1}$ .

$y = - 1 \cdot (e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}) + \frac{- 3}{- 1}$

Step 3.16.4.2.3.1.2

Rewrite $- 1 \cdot (e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}})$ as $- (e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}})$ .

$y = - (e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}}) + \frac{- 3}{- 1}$

Step 3.16.4.2.3.1.3

Divide $- 3$ by $- 1$ .

$y = - e^{- C} {| 1 - x^{2} |}^{\frac{1}{2}} + 3$

Step 4

Simplify the constant of integration.

$y = C {| 1 - x^{2} |}^{\frac{1}{2}} + 3$