Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Factor.

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Step 1.1.1

Factor out the greatest common factor from each group.

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Step 1.1.1.1

Group the first two terms and the last two terms.

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

Step 1.1.1.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.1.2

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

Cancel the common factor of $y + 1$ .

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Step 1.4.1.1

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

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

Step 1.4.1.2

Cancel the common factor.

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

Step 1.4.1.3

Rewrite the expression.

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

Step 1.4.2

Simplify the denominator.

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Step 1.4.2.1

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

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

Step 1.4.2.2

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

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y + 1$ .

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

Step 2.2.1.1.2

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

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

Step 2.2.1.1.3

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

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

Step 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Write the fraction using partial fraction decomposition.

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Step 2.3.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.3.1.1.1

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 $A$ .

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

Step 2.3.1.1.2

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

Cancel the common factor of $x + 2$ .

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Step 2.3.1.1.4.1

Cancel the common factor.

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

Step 2.3.1.1.4.2

Rewrite the expression.

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

Step 2.3.1.1.5

Cancel the common factor of $x - 2$ .

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Step 2.3.1.1.5.1

Cancel the common factor.

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

Step 2.3.1.1.5.2

Divide $x - 1$ by $1$ .

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

Step 2.3.1.1.6

Simplify each term.

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Step 2.3.1.1.6.1

Cancel the common factor of $x + 2$ .

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Step 2.3.1.1.6.1.1

Cancel the common factor.

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

Step 2.3.1.1.6.1.2

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

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

Step 2.3.1.1.6.2

Apply the distributive property.

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

Step 2.3.1.1.6.3

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

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

Step 2.3.1.1.6.4

Cancel the common factor of $x - 2$ .

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Step 2.3.1.1.6.4.1

Cancel the common factor.

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

Step 2.3.1.1.6.4.2

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

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

Step 2.3.1.1.6.5

Apply the distributive property.

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

Step 2.3.1.1.6.6

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

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

Step 2.3.1.1.7

Move $- 2 A$ .

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

Step 2.3.1.2

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

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Step 2.3.1.2.1

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.

$1 = A + B$

Step 2.3.1.2.2

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 2.3.1.2.3

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

$1 = A + B$

$- 1 = - 2 A + 2 B$

$1 = A + B$

$- 1 = - 2 A + 2 B$

Step 2.3.1.3

Solve the system of equations.

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Step 2.3.1.3.1

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

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Step 2.3.1.3.1.1

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

$A + B = 1$

$- 1 = - 2 A + 2 B$

Step 2.3.1.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$- 1 = - 2 A + 2 B$

$A = 1 - B$

$- 1 = - 2 A + 2 B$

Step 2.3.1.3.2

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

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Step 2.3.1.3.2.1

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

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

$A = 1 - B$

Step 2.3.1.3.2.2

Simplify the right side.

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Step 2.3.1.3.2.2.1

Simplify $- 2 (1 - B) + 2 B$ .

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Step 2.3.1.3.2.2.1.1

Simplify each term.

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Step 2.3.1.3.2.2.1.1.1

Apply the distributive property.

$- 1 = - 2 \cdot 1 - 2 (- B) + 2 B$

$A = 1 - B$

Step 2.3.1.3.2.2.1.1.2

Multiply $- 2$ by $1$ .

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

$A = 1 - B$

Step 2.3.1.3.2.2.1.1.3

Multiply $- 1$ by $- 2$ .

$- 1 = - 2 + 2 B + 2 B$

$A = 1 - B$

$- 1 = - 2 + 2 B + 2 B$

$A = 1 - B$

Step 2.3.1.3.2.2.1.2

Add $2 B$ and $2 B$ .

$- 1 = - 2 + 4 B$

$A = 1 - B$

$- 1 = - 2 + 4 B$

$A = 1 - B$

$- 1 = - 2 + 4 B$

$A = 1 - B$

$- 1 = - 2 + 4 B$

$A = 1 - B$

Step 2.3.1.3.3

Solve for $B$ in $- 1 = - 2 + 4 B$ .

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Step 2.3.1.3.3.1

Rewrite the equation as $- 2 + 4 B = - 1$ .

$- 2 + 4 B = - 1$

$A = 1 - B$

Step 2.3.1.3.3.2

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

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Step 2.3.1.3.3.2.1

Add $2$ to both sides of the equation.

$4 B = - 1 + 2$

$A = 1 - B$

Step 2.3.1.3.3.2.2

Add $- 1$ and $2$ .

$4 B = 1$

$A = 1 - B$

$4 B = 1$

$A = 1 - B$

Step 2.3.1.3.3.3

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

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Step 2.3.1.3.3.3.1

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

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

$A = 1 - B$

Step 2.3.1.3.3.3.2

Simplify the left side.

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Step 2.3.1.3.3.3.2.1

Cancel the common factor of $4$ .

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Step 2.3.1.3.3.3.2.1.1

Cancel the common factor.

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

$A = 1 - B$

Step 2.3.1.3.3.3.2.1.2

Divide $B$ by $1$ .

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

Step 2.3.1.3.4

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

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Step 2.3.1.3.4.1

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

$A = 1 - (\frac{1}{4})$

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

Step 2.3.1.3.4.2

Simplify the right side.

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Step 2.3.1.3.4.2.1

Simplify $1 - (\frac{1}{4})$ .

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Step 2.3.1.3.4.2.1.1

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

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

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

Step 2.3.1.3.4.2.1.2

Combine the numerators over the common denominator.

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

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

Step 2.3.1.3.4.2.1.3

Subtract $1$ from $4$ .

$A = \frac{3}{4}$

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

$A = \frac{3}{4}$

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

$A = \frac{3}{4}$

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

$A = \frac{3}{4}$

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

Step 2.3.1.3.5

List all of the solutions.

$A = \frac{3}{4}, B = \frac{1}{4}$

Step 2.3.1.4

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

$\frac{\frac{3}{4}}{x + 2} + \frac{\frac{1}{4}}{x - 2}$

Step 2.3.1.5

Simplify.

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Step 2.3.1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{3}{4} \cdot \frac{1}{x + 2} + \frac{\frac{1}{4}}{x - 2}$

Step 2.3.1.5.2

Multiply $\frac{3}{4}$ by $\frac{1}{x + 2}$ .

$\frac{3}{4 (x + 2)} + \frac{\frac{1}{4}}{x - 2}$

Step 2.3.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{3}{4 (x + 2)} + \frac{1}{4} \cdot \frac{1}{x - 2}$

Step 2.3.1.5.4

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

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

Step 2.3.2

Split the single integral into multiple integrals.

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

Step 2.3.3

Since $\frac{3}{4}$ is constant with respect to $x$ , move $\frac{3}{4}$ out of the integral.

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

Step 2.3.4

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

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Step 2.3.4.1

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

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Step 2.3.4.1.1

Differentiate $x + 2$ .

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

Step 2.3.4.1.2

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

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

Step 2.3.4.1.3

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

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

Step 2.3.4.1.4

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

$1 + 0$

Step 2.3.4.1.5

Add $1$ and $0$ .

$1$

Step 2.3.4.2

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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Step 2.3.7.1

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

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Step 2.3.7.1.1

Differentiate $x - 2$ .

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

Step 2.3.7.1.2

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

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

Step 2.3.7.1.3

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

$1 + \frac{d}{d x} [- 2]$

Step 2.3.7.1.4

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

$1 + 0$

Step 2.3.7.1.5

Add $1$ and $0$ .

$1$

Step 2.3.7.2

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

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

Step 2.3.8

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

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

Step 2.3.9

Simplify.

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

Step 2.3.10

Substitute back in for each integration substitution variable.

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Step 2.3.10.1

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

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

Step 2.3.10.2

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

Combine $\frac{3}{4}$ and $\ln (| x + 2 |)$ .

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

Step 3.1.1.2

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

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

Step 3.2

Multiply each term in $\ln (| y + 1 |) = \frac{3 \ln (| x + 2 |)}{4} + \frac{\ln (| x - 2 |)}{4} + C$ by $4$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\ln (| y + 1 |) = \frac{3 \ln (| x + 2 |)}{4} + \frac{\ln (| x - 2 |)}{4} + C$ by $4$ .

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

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Move $4$ to the left of $\ln (| y + 1 |)$ .

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

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Cancel the common factor of $4$ .

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Step 3.2.3.1.1.1

Cancel the common factor.

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

Step 3.2.3.1.1.2

Rewrite the expression.

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

Step 3.2.3.1.2

Cancel the common factor of $4$ .

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Step 3.2.3.1.2.1

Cancel the common factor.

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

Step 3.2.3.1.2.2

Rewrite the expression.

$4 \ln (| y + 1 |) = 3 \ln (| x + 2 |) + \ln (| x - 2 |) + C \cdot 4$

Step 3.2.3.1.3

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

$4 \ln (| y + 1 |) = 3 \ln (| x + 2 |) + \ln (| x - 2 |) + 4 C$

Step 3.3

Simplify the left side.

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Step 3.3.1

Simplify $4 \ln (| y + 1 |)$ .

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Step 3.3.1.1

Simplify $4 \ln (| y + 1 |)$ by moving $4$ inside the logarithm.

$\ln ({| y + 1 |}^{4}) = 3 \ln (| x + 2 |) + \ln (| x - 2 |) + 4 C$

Step 3.3.1.2

Remove the absolute value in ${| y + 1 |}^{4}$ because exponentiations with even powers are always positive.

$\ln ({(y + 1)}^{4}) = 3 \ln (| x + 2 |) + \ln (| x - 2 |) + 4 C$

Step 3.4

Simplify the right side.

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Step 3.4.1

Simplify $3 \ln (| x + 2 |) + \ln (| x - 2 |) + 4 C$ .

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Step 3.4.1.1

Simplify $3 \ln (| x + 2 |)$ by moving $3$ inside the logarithm.

$\ln ({(y + 1)}^{4}) = \ln ({| x + 2 |}^{3}) + \ln (| x - 2 |) + 4 C$

Step 3.4.1.2

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

$\ln ({(y + 1)}^{4}) = \ln ({| x + 2 |}^{3} | x - 2 |) + 4 C$

Step 3.5

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

$\ln ({(y + 1)}^{4}) - \ln ({| x + 2 |}^{3} | x - 2 |) = 4 C$

Step 3.6

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

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

Step 3.7

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

$e^{\ln (\frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |})} = e^{4 C}$

Step 3.8

Rewrite $\ln (\frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |}) = 4 C$ 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 $b^{y} = x$ .

$e^{4 C} = \frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |}$

Step 3.9

Solve for $y$ .

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Step 3.9.1

Rewrite the equation as $\frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |} = e^{4 C}$ .

$\frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |} = e^{4 C}$

Step 3.9.2

Multiply both sides by ${| x + 2 |}^{3} | x - 2 |$ .

$\frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |} ({| x + 2 |}^{3} | x - 2 |) = e^{4 C} ({| x + 2 |}^{3} | x - 2 |)$

Step 3.9.3

Simplify.

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Step 3.9.3.1

Simplify the left side.

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Step 3.9.3.1.1

Cancel the common factor of ${| x + 2 |}^{3} | x - 2 |$ .

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Step 3.9.3.1.1.1

Cancel the common factor.

$\frac{{(y + 1)}^{4}}{{| x + 2 |}^{3} | x - 2 |} ({| x + 2 |}^{3} | x - 2 |) = e^{4 C} ({| x + 2 |}^{3} | x - 2 |)$

Step 3.9.3.1.1.2

Rewrite the expression.

${(y + 1)}^{4} = e^{4 C} ({| x + 2 |}^{3} | x - 2 |)$

Step 3.9.3.2

Simplify the right side.

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Step 3.9.3.2.1

Remove parentheses.

${(y + 1)}^{4} = e^{4 C} {| x + 2 |}^{3} | x - 2 |$

Step 3.9.4

Solve for $y$ .

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Step 3.9.4.1

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

$y + 1 = \pm \sqrt[4]{e^{4 C} {| x + 2 |}^{3} | x - 2 |}$

Step 3.9.4.2

Simplify $\pm \sqrt[4]{e^{4 C} {| x + 2 |}^{3} | x - 2 |}$ .

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Step 3.9.4.2.1

Rewrite $e^{4 C} {| x + 2 |}^{3} | x - 2 |$ as ${(e^{C})}^{4} ({| x + 2 |}^{3} | x - 2 |)$ .

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Step 3.9.4.2.1.1

Rewrite $e^{4 C}$ as ${(e^{C})}^{4}$ .

$y + 1 = \pm \sqrt[4]{{(e^{C})}^{4} {| x + 2 |}^{3} | x - 2 |}$

Step 3.9.4.2.1.2

Add parentheses.

$y + 1 = \pm \sqrt[4]{{(e^{C})}^{4} ({| x + 2 |}^{3} | x - 2 |)}$

Step 3.9.4.2.2

Pull terms out from under the radical.

$y + 1 = \pm e^{C} \sqrt[4]{{| x + 2 |}^{3} | x - 2 |}$

Step 3.9.4.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.9.4.3.1

First, use the positive value of the $\pm$ to find the first solution.

$y + 1 = e^{C} \sqrt[4]{{| x + 2 |}^{3} | x - 2 |}$

Step 3.9.4.3.2

Reorder factors in $e^{C} \sqrt[4]{{| x + 2 |}^{3} | x - 2 |}$ .

$y + 1 = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C}$

Step 3.9.4.3.3

Subtract $1$ from both sides of the equation.

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

Step 3.9.4.3.4

Next, use the negative value of the $\pm$ to find the second solution.

$y + 1 = - e^{C} \sqrt[4]{{| x + 2 |}^{3} | x - 2 |}$

Step 3.9.4.3.5

Reorder factors in $- e^{C} \sqrt[4]{{| x + 2 |}^{3} | x - 2 |}$ .

$y + 1 = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C}$

Step 3.9.4.3.6

Subtract $1$ from both sides of the equation.

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

Step 3.9.4.3.7

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} e^{C} - 1$

Step 4

Simplify the constant of integration.

$y = \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} C - 1$

$y = - \sqrt[4]{{| x + 2 |}^{3} | x - 2 |} C - 1$