Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

Simplify the denominator.

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Step 1.1.1.1

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

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

Step 1.1.1.2

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 $b = y$ .

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

Step 1.1.2

Subtract $\frac{x}{(1 + y) (1 - y)}$ from both sides of the equation.

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

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

Step 1.3.2

Cancel the common factor of $x$ .

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Step 1.3.2.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Step 1.3.2.2

Cancel the common factor.

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

Step 1.3.2.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Write the fraction using partial fraction decomposition.

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Step 2.3.2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.3.2.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{1 + y}$

Step 2.3.2.1.2

$\frac{A}{1 + y} + \frac{B}{1 - y}$

Step 2.3.2.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(1 + y) (1 - y)$ .

$\frac{1 (1 + y) (1 - y)}{(1 + y) (1 - y)} = \frac{(A) (1 + y) (1 - y)}{1 + y} + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.4

Cancel the common factor of $1 + y$ .

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Step 2.3.2.1.4.1

Cancel the common factor.

$\frac{1 (1 + y) (1 - y)}{(1 + y) (1 - y)} = \frac{(A) (1 + y) (1 - y)}{1 + y} + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.4.2

Rewrite the expression.

$\frac{1 (1 - y)}{1 - y} = \frac{(A) (1 + y) (1 - y)}{1 + y} + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.5

Cancel the common factor of $1 - y$ .

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Step 2.3.2.1.5.1

Cancel the common factor.

$\frac{1 (1 - y)}{1 - y} = \frac{(A) (1 + y) (1 - y)}{1 + y} + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.5.2

Rewrite the expression.

$1 = \frac{(A) (1 + y) (1 - y)}{1 + y} + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6

Simplify each term.

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Step 2.3.2.1.6.1

Cancel the common factor of $1 + y$ .

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Step 2.3.2.1.6.1.1

Cancel the common factor.

$1 = \frac{A (1 + y) (1 - y)}{1 + y} + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6.1.2

Divide $(A) (1 - y)$ by $1$ .

$1 = (A) (1 - y) + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6.2

Apply the distributive property.

$1 = A \cdot 1 + A (- y) + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6.3

Multiply $A$ by $1$ .

$1 = A + A (- y) + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6.4

Rewrite using the commutative property of multiplication.

$1 = A - A y + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6.5

Cancel the common factor of $1 - y$ .

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Step 2.3.2.1.6.5.1

Cancel the common factor.

$1 = A - A y + \frac{(B) (1 + y) (1 - y)}{1 - y}$

Step 2.3.2.1.6.5.2

Divide $(B) (1 + y)$ by $1$ .

$1 = A - A y + (B) (1 + y)$

Step 2.3.2.1.6.6

Apply the distributive property.

$1 = A - A y + B \cdot 1 + B y$

Step 2.3.2.1.6.7

Multiply $B$ by $1$ .

$1 = A - A y + B + B y$

Step 2.3.2.1.7

Simplify the expression.

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Step 2.3.2.1.7.1

Move $A$ .

$1 = A - y A + B + B y$

Step 2.3.2.1.7.2

Reorder $B$ and $y$ .

$1 = A - y A + B + B y$

Step 2.3.2.1.7.3

Move $B$ .

$1 = A - y A + B y + B$

Step 2.3.2.1.7.4

Move $A$ .

$1 = - y A + B y + A + B$

Step 2.3.2.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.3.2.2.1

Create an equation for the partial fraction variables by equating the coefficients of $y$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = - 1 A + B$

Step 2.3.2.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $y$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A + B$

Step 2.3.2.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = - 1 A + B$

$1 = A + B$

$0 = - 1 A + B$

$1 = A + B$

Step 2.3.2.3

Solve the system of equations.

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Step 2.3.2.3.1

Solve for $A$ in $1 = A + B$ .

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Step 2.3.2.3.1.1

Rewrite the equation as $A + B = 1$ .

$A + B = 1$

$0 = - 1 A + B$

Step 2.3.2.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$0 = - 1 A + B$

$A = 1 - B$

$0 = - 1 A + B$

Step 2.3.2.3.2

Replace all occurrences of $A$ with $1 - B$ in each equation.

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Step 2.3.2.3.2.1

Replace all occurrences of $A$ in $0 = - 1 A + B$ with $1 - B$ .

$0 = - 1 (1 - B) + B$

$A = 1 - B$

Step 2.3.2.3.2.2

Simplify the right side.

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Step 2.3.2.3.2.2.1

Simplify $- 1 (1 - B) + B$ .

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Step 2.3.2.3.2.2.1.1

Simplify each term.

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Step 2.3.2.3.2.2.1.1.1

Apply the distributive property.

$0 = - 1 \cdot 1 - 1 (- B) + B$

$A = 1 - B$

Step 2.3.2.3.2.2.1.1.2

Multiply $- 1$ by $1$ .

$0 = - 1 - 1 (- B) + B$

$A = 1 - B$

Step 2.3.2.3.2.2.1.1.3

Multiply $- 1 (- B)$ .

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Step 2.3.2.3.2.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$0 = - 1 + 1 B + B$

$A = 1 - B$

Step 2.3.2.3.2.2.1.1.3.2

Multiply $B$ by $1$ .

$0 = - 1 + B + B$

$A = 1 - B$

$0 = - 1 + B + B$

$A = 1 - B$

$0 = - 1 + B + B$

$A = 1 - B$

Step 2.3.2.3.2.2.1.2

Add $B$ and $B$ .

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

$0 = - 1 + 2 B$

$A = 1 - B$

Step 2.3.2.3.3

Solve for $B$ in $0 = - 1 + 2 B$ .

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Step 2.3.2.3.3.1

Rewrite the equation as $- 1 + 2 B = 0$ .

$- 1 + 2 B = 0$

$A = 1 - B$

Step 2.3.2.3.3.2

Add $1$ to both sides of the equation.

$2 B = 1$

$A = 1 - B$

Step 2.3.2.3.3.3

Divide each term in $2 B = 1$ by $2$ and simplify.

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Step 2.3.2.3.3.3.1

Divide each term in $2 B = 1$ by $2$ .

$\frac{2 B}{2} = \frac{1}{2}$

$A = 1 - B$

Step 2.3.2.3.3.3.2

Simplify the left side.

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Step 2.3.2.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{1}{2}$

$A = 1 - B$

Step 2.3.2.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

$B = \frac{1}{2}$

$A = 1 - B$

Step 2.3.2.3.4

Replace all occurrences of $B$ with $\frac{1}{2}$ in each equation.

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Step 2.3.2.3.4.1

Replace all occurrences of $B$ in $A = 1 - B$ with $\frac{1}{2}$ .

$A = 1 - (\frac{1}{2})$

$B = \frac{1}{2}$

Step 2.3.2.3.4.2

Simplify the right side.

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Step 2.3.2.3.4.2.1

Simplify $1 - (\frac{1}{2})$ .

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Step 2.3.2.3.4.2.1.1

Write $1$ as a fraction with a common denominator.

$A = \frac{2}{2} - \frac{1}{2}$

$B = \frac{1}{2}$

Step 2.3.2.3.4.2.1.2

Combine the numerators over the common denominator.

$A = \frac{2 - 1}{2}$

$B = \frac{1}{2}$

Step 2.3.2.3.4.2.1.3

Subtract $1$ from $2$ .

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

$A = \frac{1}{2}$

$B = \frac{1}{2}$

Step 2.3.2.3.5

List all of the solutions.

$A = \frac{1}{2}, B = \frac{1}{2}$

Step 2.3.2.4

Replace each of the partial fraction coefficients in $\frac{A}{1 + y} + \frac{B}{1 - y}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{2}}{1 + y} + \frac{\frac{1}{2}}{1 - y}$

Step 2.3.2.5

Simplify.

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Step 2.3.2.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2} \cdot \frac{1}{1 + y} + \frac{\frac{1}{2}}{1 - y}$

Step 2.3.2.5.2

Multiply $\frac{1}{2}$ by $\frac{1}{1 + y}$ .

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

Step 2.3.2.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 (1 + y)} + \frac{1}{2} \cdot \frac{1}{1 - y}$

Step 2.3.2.5.4

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

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

Step 2.3.3

Split the single integral into multiple integrals.

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

Step 2.3.4

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

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

Step 2.3.5

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

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Step 2.3.5.1

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

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Step 2.3.5.1.1

Differentiate $1 + y$ .

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

Step 2.3.5.1.2

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

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

Step 2.3.5.1.3

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

$0 + \frac{d}{d y} [y]$

Step 2.3.5.1.4

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

$0 + 1$

Step 2.3.5.1.5

Add $0$ and $1$ .

$1$

Step 2.3.5.2

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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

Step 2.3.8

Let $u_{2} = 1 - y$ . Then $d u_{2} = - d y$ , so $- d u_{2} = d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.8.1

Let $u_{2} = 1 - y$ . Find $\frac{d u_{2}}{d y}$ .

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Step 2.3.8.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.8.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.8.2

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

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

Step 2.3.9

Move the negative in front of the fraction.

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

Step 2.3.10

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

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

Step 2.3.11

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

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

Step 2.3.12

Simplify.

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

Step 2.3.13

Substitute back in for each integration substitution variable.

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Step 2.3.13.1

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

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

Step 2.3.13.2

Replace all occurrences of $u_{2}$ with $1 - y$ .

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

Step 2.3.14

Simplify.

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Step 2.3.14.1

Combine the numerators over the common denominator.

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

Step 2.3.14.2

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

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

Step 2.3.15

Reorder terms.

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

Step 2.4

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

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

Step 3

Solve for $x$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Simplify terms.

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Step 3.4.1

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

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

Step 3.4.2

Combine the numerators over the common denominator.

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

Step 3.5

Move $2$ to the left of $\ln (| x |)$ .

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

Step 3.6

Simplify the left side.

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Step 3.6.1

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

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Step 3.6.1.1

Simplify the numerator.

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Step 3.6.1.1.1

Simplify $2 \ln (| x |)$ by moving $2$ inside the logarithm.

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

Step 3.6.1.1.2

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

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

Step 3.6.1.1.3

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

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

Step 3.6.1.1.4

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

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

Step 3.6.1.2

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

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

Step 3.6.1.3

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

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

Step 3.6.1.4

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.6.1.4.1

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

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

Step 3.6.1.4.2

Apply the product rule to $x^{2} | 1 + y |$ .

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

Step 3.6.1.5

Simplify the numerator.

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Step 3.6.1.5.1

Multiply the exponents in ${(x^{2})}^{\frac{1}{2}}$ .

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Step 3.6.1.5.1.1

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

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

Step 3.6.1.5.1.2

Cancel the common factor of $2$ .

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Step 3.6.1.5.1.2.1

Cancel the common factor.

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

Step 3.6.1.5.1.2.2

Rewrite the expression.

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

Step 3.6.1.5.2

Simplify.

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

Solve for $x$ .

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Step 3.9.1

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

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

Step 3.9.2

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

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

Step 3.9.3

Simplify the left side.

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Step 3.9.3.1

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

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Step 3.9.3.1.1

Cancel the common factor.

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

Step 3.9.3.1.2

Rewrite the expression.

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

Step 3.9.4

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

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Step 3.9.4.1

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

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

Step 3.9.4.2

Simplify the left side.

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Step 3.9.4.2.1

Cancel the common factor.

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

Step 3.9.4.2.2

Divide $x$ by $1$ .

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

Step 4

Simplify the constant of integration.

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