Calculus Examples

$(1 + 2 y) d x + (4 - x^{2}) d y = 0$

Step 1

Subtract $(1 + 2 y) d x$ from both sides of the equation.

$(4 - x^{2}) d y = - (1 + 2 y) d x$

Step 2

Multiply both sides by $\frac{1}{(4 - x^{2}) (1 + 2 y)}$ .

$\frac{1}{(4 - x^{2}) (1 + 2 y)} (4 - x^{2}) d y = \frac{1}{(4 - x^{2}) (1 + 2 y)} (- (1 + 2 y)) d x$

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $4 - x^{2}$ .

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Step 3.1.1

Cancel the common factor.

$\frac{1}{(4 - x^{2}) (1 + 2 y)} (4 - x^{2}) d y = \frac{1}{(4 - x^{2}) (1 + 2 y)} (- (1 + 2 y)) d x$

Step 3.1.2

Rewrite the expression.

$\frac{1}{1 + 2 y} d y = \frac{1}{(4 - x^{2}) (1 + 2 y)} (- (1 + 2 y)) d x$

Step 3.2

Rewrite using the commutative property of multiplication.

$\frac{1}{1 + 2 y} d y = - \frac{1}{(4 - x^{2}) (1 + 2 y)} (1 + 2 y) d x$

Step 3.3

Cancel the common factor of $1 + 2 y$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{(4 - x^{2}) (1 + 2 y)}$ into the numerator.

$\frac{1}{1 + 2 y} d y = \frac{- 1}{(4 - x^{2}) (1 + 2 y)} (1 + 2 y) d x$

Step 3.3.2

Factor $1 + 2 y$ out of $(4 - x^{2}) (1 + 2 y)$ .

$\frac{1}{1 + 2 y} d y = \frac{- 1}{(1 + 2 y) (4 - x^{2})} (1 + 2 y) d x$

Step 3.3.3

Cancel the common factor.

$\frac{1}{1 + 2 y} d y = \frac{- 1}{(1 + 2 y) (4 - x^{2})} (1 + 2 y) d x$

Step 3.3.4

Rewrite the expression.

$\frac{1}{1 + 2 y} d y = \frac{- 1}{4 - x^{2}} d x$

Step 3.4

Simplify the denominator.

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Step 3.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 3.4.2

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

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

Step 3.5

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $1 + 2 y$ .

$\frac{d}{d y} [1 + 2 y]$

Step 4.2.1.1.2

Differentiate.

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Step 4.2.1.1.2.1

By the Sum Rule, the derivative of $1 + 2 y$ with respect to $y$ is $\frac{d}{d y} [1] + \frac{d}{d y} [2 y]$ .

$\frac{d}{d y} [1] + \frac{d}{d y} [2 y]$

Step 4.2.1.1.2.2

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

$0 + \frac{d}{d y} [2 y]$

Step 4.2.1.1.3

Evaluate $\frac{d}{d y} [2 y]$ .

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Step 4.2.1.1.3.1

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

$0 + 2 \frac{d}{d y} [y]$

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 $n = 1$ .

$0 + 2 \cdot 1$

Step 4.2.1.1.3.3

Multiply $2$ by $1$ .

$0 + 2$

Step 4.2.1.1.4

Add $0$ and $2$ .

$2$

Step 4.2.1.2

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

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

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

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

Step 4.2.2.2

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

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

Step 4.2.3

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

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

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

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.

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

Step 4.3.2

Write the fraction using partial fraction decomposition.

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Step 4.3.2.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.3.2.1.1

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 $A$ .

$\frac{A}{2 + x}$

Step 4.3.2.1.2

$\frac{A}{2 + x} + \frac{B}{2 - x}$

Step 4.3.2.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(2 + x) (2 - x)$ .

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

Step 4.3.2.1.4

Cancel the common factor of $2 + x$ .

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Step 4.3.2.1.4.1

Cancel the common factor.

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

Step 4.3.2.1.4.2

Rewrite the expression.

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

Step 4.3.2.1.5

Cancel the common factor of $2 - x$ .

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Step 4.3.2.1.5.1

Cancel the common factor.

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

Step 4.3.2.1.5.2

Rewrite the expression.

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

Step 4.3.2.1.6

Simplify each term.

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Step 4.3.2.1.6.1

Cancel the common factor of $2 + x$ .

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Step 4.3.2.1.6.1.1

Cancel the common factor.

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

Step 4.3.2.1.6.1.2

Divide $(A) (2 - x)$ by $1$ .

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

Step 4.3.2.1.6.2

Apply the distributive property.

$1 = A \cdot 2 + A (- x) + \frac{(B) (2 + x) (2 - x)}{2 - x}$

Step 4.3.2.1.6.3

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

$1 = 2 \cdot A + A (- x) + \frac{(B) (2 + x) (2 - x)}{2 - x}$

Step 4.3.2.1.6.4

Rewrite using the commutative property of multiplication.

$1 = 2 \cdot A - A x + \frac{(B) (2 + x) (2 - x)}{2 - x}$

Step 4.3.2.1.6.5

Cancel the common factor of $2 - x$ .

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Step 4.3.2.1.6.5.1

Cancel the common factor.

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

Step 4.3.2.1.6.5.2

Divide $(B) (2 + x)$ by $1$ .

$1 = 2 A - A x + (B) (2 + x)$

Step 4.3.2.1.6.6

Apply the distributive property.

$1 = 2 A - A x + B \cdot 2 + B x$

Step 4.3.2.1.6.7

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

$1 = 2 A - A x + 2 B + B x$

Step 4.3.2.1.7

Simplify the expression.

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Step 4.3.2.1.7.1

Move $A$ .

$1 = 2 A - x A + 2 B + B x$

Step 4.3.2.1.7.2

Reorder $B$ and $x$ .

$1 = 2 A - x A + 2 B + B x$

Step 4.3.2.1.7.3

Move $2 B$ .

$1 = 2 A - x A + B x + 2 B$

Step 4.3.2.1.7.4

Move $2 A$ .

$1 = - x A + B x + 2 A + 2 B$

Step 4.3.2.2

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

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Step 4.3.2.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ 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 4.3.2.2.2

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

$1 = 2 A + 2 B$

Step 4.3.2.2.3

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

$0 = - 1 A + B$

$1 = 2 A + 2 B$

$0 = - 1 A + B$

$1 = 2 A + 2 B$

Step 4.3.2.3

Solve the system of equations.

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Step 4.3.2.3.1

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

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Step 4.3.2.3.1.1

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

$- 1 A + B = 0$

$1 = 2 A + 2 B$

Step 4.3.2.3.1.2

Rewrite $- 1 A$ as $- A$ .

$- A + B = 0$

$1 = 2 A + 2 B$

Step 4.3.2.3.1.3

Add $A$ to both sides of the equation.

$B = A$

$1 = 2 A + 2 B$

$B = A$

$1 = 2 A + 2 B$

Step 4.3.2.3.2

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

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Step 4.3.2.3.2.1

Replace all occurrences of $B$ in $1 = 2 A + 2 B$ with $A$ .

$1 = 2 A + 2 (A)$

$B = A$

Step 4.3.2.3.2.2

Simplify the right side.

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Step 4.3.2.3.2.2.1

Add $2 A$ and $2 A$ .

$1 = 4 A$

$B = A$

$1 = 4 A$

$B = A$

$1 = 4 A$

$B = A$

Step 4.3.2.3.3

Solve for $A$ in $1 = 4 A$ .

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Step 4.3.2.3.3.1

Rewrite the equation as $4 A = 1$ .

$4 A = 1$

$B = A$

Step 4.3.2.3.3.2

Divide each term in $4 A = 1$ by $4$ and simplify.

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Step 4.3.2.3.3.2.1

Divide each term in $4 A = 1$ by $4$ .

$\frac{4 A}{4} = \frac{1}{4}$

$B = A$

Step 4.3.2.3.3.2.2

Simplify the left side.

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Step 4.3.2.3.3.2.2.1

Cancel the common factor of $4$ .

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Step 4.3.2.3.3.2.2.1.1

Cancel the common factor.

$\frac{4 A}{4} = \frac{1}{4}$

$B = A$

Step 4.3.2.3.3.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{4}$

$B = A$

$A = \frac{1}{4}$

$B = A$

$A = \frac{1}{4}$

$B = A$

$A = \frac{1}{4}$

$B = A$

$A = \frac{1}{4}$

$B = A$

Step 4.3.2.3.4

Replace all occurrences of $A$ with $\frac{1}{4}$ in each equation.

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Step 4.3.2.3.4.1

Replace all occurrences of $A$ in $B = A$ with $\frac{1}{4}$ .

$B = \frac{1}{4}$

$A = \frac{1}{4}$

Step 4.3.2.3.4.2

Simplify the left side.

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Step 4.3.2.3.4.2.1

Remove parentheses.

$B = \frac{1}{4}$

$A = \frac{1}{4}$

$B = \frac{1}{4}$

$A = \frac{1}{4}$

$B = \frac{1}{4}$

$A = \frac{1}{4}$

Step 4.3.2.3.5

List all of the solutions.

$B = \frac{1}{4}, A = \frac{1}{4}$

Step 4.3.2.4

Replace each of the partial fraction coefficients in $\frac{A}{2 + x} + \frac{B}{2 - x}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{4}}{2 + x} + \frac{\frac{1}{4}}{2 - x}$

Step 4.3.2.5

Simplify.

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Step 4.3.2.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{2 + x} + \frac{\frac{1}{4}}{2 - x}$

Step 4.3.2.5.2

Multiply $\frac{1}{4}$ by $\frac{1}{2 + x}$ .

$\frac{1}{4 (2 + x)} + \frac{\frac{1}{4}}{2 - x}$

Step 4.3.2.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4 (2 + x)} + \frac{1}{4} \cdot \frac{1}{2 - x}$

Step 4.3.2.5.4

Multiply $\frac{1}{4}$ by $\frac{1}{2 - x}$ .

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

Step 4.3.3

Split the single integral into multiple integrals.

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

Step 4.3.4

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

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

Step 4.3.5

Let $u_{2} = 2 + x$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 4.3.5.1

Let $u_{2} = 2 + x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 4.3.5.1.1

Differentiate $2 + x$ .

$\frac{d}{d x} [2 + x]$

Step 4.3.5.1.2

By the Sum Rule, the derivative of $2 + x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [x]$ .

$\frac{d}{d x} [2] + \frac{d}{d x} [x]$

Step 4.3.5.1.3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 4.3.5.1.4

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

$0 + 1$

Step 4.3.5.1.5

Add $0$ and $1$ .

$1$

Step 4.3.5.2

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

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

Step 4.3.6

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

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

Step 4.3.7

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

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

Step 4.3.8

Let $u_{3} = 2 - x$ . Then $d u_{3} = - d x$ , so $- d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 4.3.8.1

Let $u_{3} = 2 - x$ . Find $\frac{d u_{3}}{d x}$ .

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Step 4.3.8.1.1

Rewrite.

$\frac{- 1}{1}$

Step 4.3.8.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 4.3.8.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

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

Step 4.3.9

Move the negative in front of the fraction.

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

Step 4.3.10

Since $- 1$ is constant with respect to $u_{3}$ , move $- 1$ out of the integral.

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

Step 4.3.11

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

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

Step 4.3.12

Simplify.

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

Step 4.3.13

Substitute back in for each integration substitution variable.

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Step 4.3.13.1

Replace all occurrences of $u_{2}$ with $2 + x$ .

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

Step 4.3.13.2

Replace all occurrences of $u_{3}$ with $2 - x$ .

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

Step 4.3.14

Simplify.

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Step 4.3.14.1

Combine the numerators over the common denominator.

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

Step 4.3.14.2

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

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

Step 4.3.15

Reorder terms.

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

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

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Step 5.2.1.1.1

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

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

Step 5.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $2 (- \frac{1}{4} \ln (\frac{| 2 + x |}{| 2 - x |}) + C)$ .

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Step 5.2.2.1.1

Combine $\ln (\frac{| 2 + x |}{| 2 - x |})$ and $\frac{1}{4}$ .

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

Step 5.2.2.1.2

To write $C$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 5.2.2.1.3

Simplify terms.

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Step 5.2.2.1.3.1

Combine $C$ and $\frac{4}{4}$ .

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

Step 5.2.2.1.3.2

Combine the numerators over the common denominator.

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

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 $4$ .

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

Step 5.2.2.1.3.3.2

Cancel the common factor.

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

Step 5.2.2.1.3.3.3

Rewrite the expression.

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

Step 5.2.2.1.4

Move $4$ to the left of $C$ .

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

Step 5.2.2.1.5

Simplify with factoring out.

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Step 5.2.2.1.5.1

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

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

Step 5.2.2.1.5.2

Factor $- 1$ out of $4 C$ .

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

Step 5.2.2.1.5.3

Factor $- 1$ out of $- (\ln (\frac{| 2 + x |}{| 2 - x |})) - (- 4 C)$ .

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

Step 5.2.2.1.5.4

Simplify the expression.

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Step 5.2.2.1.5.4.1

Rewrite $- (\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C)$ as $- 1 (\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C)$ .

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

Step 5.2.2.1.5.4.2

Move the negative in front of the fraction.

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

Step 5.3

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

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

Step 5.4

Rewrite $\ln (| 1 + 2 y |) = - \frac{\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C}{2}$ 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 $b^{y} = x$ .

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

Step 5.5

Solve for $y$ .

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Step 5.5.1

Rewrite the equation as $| 1 + 2 y | = e^{- \frac{\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C}{2}}$ .

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

Step 5.5.2

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

$1 + 2 y = \pm e^{- \frac{\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C}{2}}$

Step 5.5.3

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

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Step 5.5.3.1

Subtract $1$ from both sides of the equation.

$2 y = \pm e^{- \frac{\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C}{2}} - 1$

Step 5.5.3.2

Simplify each term.

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Step 5.5.3.2.1

Split the fraction $\frac{\ln (\frac{| 2 + x |}{| 2 - x |}) - 4 C}{2}$ into two fractions.

$2 y = \pm e^{- (\frac{\ln (\frac{| 2 + x |}{| 2 - x |})}{2} + \frac{- 4 C}{2})} - 1$

Step 5.5.3.2.2

Simplify each term.

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Step 5.5.3.2.2.1

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

$2 y = \pm e^{- (\frac{1}{2} \ln (\frac{| 2 + x |}{| 2 - x |}) + \frac{- 4 C}{2})} - 1$

Step 5.5.3.2.2.2

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

$2 y = \pm e^{- (\ln ({(\frac{| 2 + x |}{| 2 - x |})}^{\frac{1}{2}}) + \frac{- 4 C}{2})} - 1$

Step 5.5.3.2.2.3

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

$2 y = \pm e^{- (\ln (\frac{{| 2 + x |}^{\frac{1}{2}}}{{| 2 - x |}^{\frac{1}{2}}}) + \frac{- 4 C}{2})} - 1$

Step 5.5.3.2.2.4

Cancel the common factor of $- 4$ and $2$ .

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Step 5.5.3.2.2.4.1

Factor $2$ out of $- 4 C$ .

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

Step 5.5.3.2.2.4.2

Cancel the common factors.

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Step 5.5.3.2.2.4.2.1

Factor $2$ out of $2$ .

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

Step 5.5.3.2.2.4.2.2

Cancel the common factor.

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

Step 5.5.3.2.2.4.2.3

Rewrite the expression.

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

Step 5.5.3.2.2.4.2.4

Divide $- 2 C$ by $1$ .

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

Step 5.5.3.2.3

Apply the distributive property.

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

Step 5.5.3.2.4

Multiply $- 2$ by $- 1$ .

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

Step 5.5.4

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

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Step 5.5.4.1

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

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

Step 5.5.4.2

Simplify the left side.

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Step 5.5.4.2.1

Cancel the common factor of $2$ .

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Step 5.5.4.2.1.1

Cancel the common factor.

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

Step 5.5.4.2.1.2

Divide $y$ by $1$ .

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

Step 5.5.4.3

Simplify the right side.

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Step 5.5.4.3.1

Move the negative in front of the fraction.

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

Step 5.5.4.3.2

Combine the numerators over the common denominator.

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

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

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

Step 6.2

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

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

Step 6.3

Reorder $e^{- \ln (\frac{{| 2 + x |}^{\frac{1}{2}}}{{| 2 - x |}^{\frac{1}{2}}})}$ and $e^{C}$ .

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

Step 6.4

Combine constants with the plus or minus.

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