Calculus Examples

$2 x y + 6 x + (x^{2} - 4) \frac{d y}{d x} = 0$

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

Apply the distributive property.

$2 x y + 6 x + x^{2} \frac{d y}{d x} - 4 \frac{d y}{d x} = 0$

Step 1.1.2

Move all terms not containing $\frac{d y}{d x}$ to the right side of the equation.

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Step 1.1.2.1

Subtract $2 x y$ from both sides of the equation.

$6 x + x^{2} \frac{d y}{d x} - 4 \frac{d y}{d x} = - 2 x y$

Step 1.1.2.2

Subtract $6 x$ from both sides of the equation.

$x^{2} \frac{d y}{d x} - 4 \frac{d y}{d x} = - 2 x y - 6 x$

Step 1.1.3

Factor $\frac{d y}{d x}$ out of $x^{2} \frac{d y}{d x} - 4 \frac{d y}{d x}$ .

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Step 1.1.3.1

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

$\frac{d y}{d x} x^{2} - 4 \frac{d y}{d x} = - 2 x y - 6 x$

Step 1.1.3.2

Factor $\frac{d y}{d x}$ out of $- 4 \frac{d y}{d x}$ .

$\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot - 4 = - 2 x y - 6 x$

Step 1.1.3.3

Factor $\frac{d y}{d x}$ out of $\frac{d y}{d x} x^{2} + \frac{d y}{d x} \cdot - 4$ .

$\frac{d y}{d x} (x^{2} - 4) = - 2 x y - 6 x$

Step 1.1.4

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

$\frac{d y}{d x} (x^{2} - 2^{2}) = - 2 x y - 6 x$

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 $a = x$ and $b = 2$ .

$\frac{d y}{d x} ((x + 2) (x - 2)) = - 2 x y - 6 x$

Step 1.1.5.2

Remove unnecessary parentheses.

$\frac{d y}{d x} (x + 2) (x - 2) = - 2 x y - 6 x$

Step 1.1.6

Divide each term in $\frac{d y}{d x} (x + 2) (x - 2) = - 2 x y - 6 x$ by $(x + 2) (x - 2)$ and simplify.

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Step 1.1.6.1

Divide each term in $\frac{d y}{d x} (x + 2) (x - 2) = - 2 x y - 6 x$ by $(x + 2) (x - 2)$ .

$\frac{\frac{d y}{d x} (x + 2) (x - 2)}{(x + 2) (x - 2)} = \frac{- 2 x y}{(x + 2) (x - 2)} + \frac{- 6 x}{(x + 2) (x - 2)}$

Step 1.1.6.2

Simplify the left side.

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Step 1.1.6.2.1

Cancel the common factor of $x + 2$ .

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Step 1.1.6.2.1.1

Cancel the common factor.

$\frac{\frac{d y}{d x} (x + 2) (x - 2)}{(x + 2) (x - 2)} = \frac{- 2 x y}{(x + 2) (x - 2)} + \frac{- 6 x}{(x + 2) (x - 2)}$

Step 1.1.6.2.1.2

Rewrite the expression.

$\frac{\frac{d y}{d x} (x - 2)}{x - 2} = \frac{- 2 x y}{(x + 2) (x - 2)} + \frac{- 6 x}{(x + 2) (x - 2)}$

Step 1.1.6.2.2

Cancel the common factor of $x - 2$ .

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Step 1.1.6.2.2.1

Cancel the common factor.

$\frac{\frac{d y}{d x} (x - 2)}{x - 2} = \frac{- 2 x y}{(x + 2) (x - 2)} + \frac{- 6 x}{(x + 2) (x - 2)}$

Step 1.1.6.2.2.2

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

$\frac{d y}{d x} = \frac{- 2 x y}{(x + 2) (x - 2)} + \frac{- 6 x}{(x + 2) (x - 2)}$

Step 1.1.6.3

Simplify the right side.

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Step 1.1.6.3.1

Simplify each term.

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Step 1.1.6.3.1.1

Move the negative in front of the fraction.

$\frac{d y}{d x} = - \frac{2 x y}{(x + 2) (x - 2)} + \frac{- 6 x}{(x + 2) (x - 2)}$

Step 1.1.6.3.1.2

Move the negative in front of the fraction.

$\frac{d y}{d x} = - \frac{2 x y}{(x + 2) (x - 2)} - \frac{6 x}{(x + 2) (x - 2)}$

Step 1.2

Factor.

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Step 1.2.1

Factor $- 1$ out of $- \frac{2 x y}{(x + 2) (x - 2)} - \frac{6 x}{(x + 2) (x - 2)}$ .

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Step 1.2.1.1

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

$\frac{d y}{d x} = - (\frac{2 x y}{(x + 2) (x - 2)}) - \frac{6 x}{(x + 2) (x - 2)}$

Step 1.2.1.2

Factor $- 1$ out of $- \frac{6 x}{(x + 2) (x - 2)}$ .

$\frac{d y}{d x} = - (\frac{2 x y}{(x + 2) (x - 2)}) - (\frac{6 x}{(x + 2) (x - 2)})$

Step 1.2.1.3

Factor $- 1$ out of $- (\frac{2 x y}{(x + 2) (x - 2)}) - (\frac{6 x}{(x + 2) (x - 2)})$ .

$\frac{d y}{d x} = - (\frac{2 x y}{(x + 2) (x - 2)} + \frac{6 x}{(x + 2) (x - 2)})$

Step 1.2.2

Combine the numerators over the common denominator.

$\frac{d y}{d x} = - \frac{2 x y + 6 x}{(x + 2) (x - 2)}$

Step 1.2.3

Factor $2 x$ out of $2 x y + 6 x$ .

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Step 1.2.3.1

Factor $2 x$ out of $2 x y$ .

$\frac{d y}{d x} = - \frac{2 x (y) + 6 x}{(x + 2) (x - 2)}$

Step 1.2.3.2

Factor $2 x$ out of $6 x$ .

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

Step 1.2.3.3

Factor $2 x$ out of $2 x (y) + 2 x (3)$ .

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

Step 1.3

Regroup factors.

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

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

Rewrite using the commutative property of multiplication.

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

Step 1.5.2

Cancel the common factor of $y + 3$ .

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Step 1.5.2.1

Move the leading negative in $- \frac{1}{y + 3}$ into the numerator.

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

Step 1.5.2.2

Factor $y + 3$ out of $\frac{2 x}{(x + 2) (x - 2)} (y + 3)$ .

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

Step 1.5.2.3

Cancel the common factor.

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

Step 1.5.2.4

Rewrite the expression.

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

Step 1.6

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u_{1} = y + 3$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y + 3$ .

$\frac{d}{d y} [y + 3]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y + 3$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [3]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [3]$

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 $n = 1$ .

$1 + \frac{d}{d y} [3]$

Step 2.2.1.1.4

Since $3$ is constant with respect to $y$ , the derivative of $3$ 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 $d u_{1}$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 2.3.3

Multiply $2$ by $- 1$ .

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

Step 2.3.4

Let $u_{2} = (x + 2) (x - 2)$ . 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 2.3.4.1

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

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Step 2.3.4.1.1

Differentiate $(x + 2) (x - 2)$ .

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

Step 2.3.4.1.2

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 $f (x) = x + 2$ and $g (x) = x - 2$ .

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

Step 2.3.4.1.3

Differentiate.

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Step 2.3.4.1.3.1

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

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

Step 2.3.4.1.3.2

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

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

Step 2.3.4.1.3.3

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

$(x + 2) (1 + 0) + (x - 2) \frac{d}{d x} [x + 2]$

Step 2.3.4.1.3.4

Simplify the expression.

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Step 2.3.4.1.3.4.1

Add $1$ and $0$ .

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

Step 2.3.4.1.3.4.2

Multiply $x + 2$ by $1$ .

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

Step 2.3.4.1.3.5

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

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

Step 2.3.4.1.3.6

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

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

Step 2.3.4.1.3.7

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

$x + 2 + (x - 2) (1 + 0)$

Step 2.3.4.1.3.8

Simplify by adding terms.

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Step 2.3.4.1.3.8.1

Add $1$ and $0$ .

$x + 2 + (x - 2) \cdot 1$

Step 2.3.4.1.3.8.2

Multiply $x - 2$ by $1$ .

$x + 2 + x - 2$

Step 2.3.4.1.3.8.3

Add $x$ and $x$ .

$2 x + 2 - 2$

Step 2.3.4.1.3.8.4

Simplify by subtracting numbers.

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Step 2.3.4.1.3.8.4.1

Subtract $2$ from $2$ .

$2 x + 0$

Step 2.3.4.1.3.8.4.2

Add $2 x$ and $0$ .

$2 x$

Step 2.3.4.2

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

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

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

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

Step 2.3.6

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

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

Step 2.3.7

Simplify.

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Step 2.3.7.1

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

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

Step 2.3.7.2

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

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Step 2.3.7.2.1

Factor $2$ out of $- 2$ .

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

Step 2.3.7.2.2

Cancel the common factors.

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Step 2.3.7.2.2.1

Factor $2$ out of $2$ .

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

Step 2.3.7.2.2.2

Cancel the common factor.

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

Step 2.3.7.2.2.3

Rewrite the expression.

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

Step 2.3.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 2.3.8

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

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

Step 2.3.9

Simplify.

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

Step 2.3.10

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Expand $(x + 2) (x - 2)$ using the FOIL Method.

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Step 3.3.1

Apply the distributive property.

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

Step 3.3.2

Apply the distributive property.

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

Step 3.3.3

Apply the distributive property.

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

Step 3.4

Simplify terms.

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Step 3.4.1

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

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Step 3.4.1.1

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

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

Step 3.4.1.2

Add $- 2 x$ and $2 x$ .

$\ln (| y + 3 | | x \cdot x + 0 + 2 \cdot - 2 |) = C$

Step 3.4.1.3

Add $x \cdot x$ and $0$ .

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

Step 3.4.2

Simplify each term.

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Step 3.4.2.1

Multiply $x$ by $x$ .

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

Step 3.4.2.2

Multiply $2$ by $- 2$ .

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

Step 3.5

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

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

Step 3.6

Expand $(y + 3) (x^{2} - 4)$ using the FOIL Method.

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Step 3.6.1

Apply the distributive property.

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

Step 3.6.2

Apply the distributive property.

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

Step 3.6.3

Apply the distributive property.

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

Step 3.7

Simplify each term.

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Step 3.7.1

Move $- 4$ to the left of $y$ .

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

Step 3.7.2

Multiply $3$ by $- 4$ .

$\ln (| y x^{2} - 4 y + 3 x^{2} - 12 |) = C$

Step 3.8

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

$e^{\ln (| y x^{2} - 4 y + 3 x^{2} - 12 |)} = e^{C}$

Step 3.9

Rewrite $\ln (| y x^{2} - 4 y + 3 x^{2} - 12 |) = C$ 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 $b^{y} = x$ .

$e^{C} = | y x^{2} - 4 y + 3 x^{2} - 12 |$

Step 3.10

Solve for $y$ .

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Step 3.10.1

Rewrite the equation as $| y x^{2} - 4 y + 3 x^{2} - 12 | = e^{C}$ .

$| y x^{2} - 4 y + 3 x^{2} - 12 | = e^{C}$

Step 3.10.2

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

$y x^{2} - 4 y + 3 x^{2} - 12 = \pm e^{C}$

Step 3.10.3

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

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Step 3.10.3.1

Subtract $3 x^{2}$ from both sides of the equation.

$y x^{2} - 4 y - 12 = \pm e^{C} - 3 x^{2}$

Step 3.10.3.2

Add $12$ to both sides of the equation.

$y x^{2} - 4 y = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.4

Factor $y$ out of $y x^{2} - 4 y$ .

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Step 3.10.4.1

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

$y (x^{2}) - 4 y = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.4.2

Factor $y$ out of $- 4 y$ .

$y (x^{2}) + y \cdot - 4 = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.4.3

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

$y (x^{2} - 4) = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.5

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

$y (x^{2} - 2^{2}) = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.6

Factor.

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Step 3.10.6.1

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

$y ((x + 2) (x - 2)) = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.6.2

Remove unnecessary parentheses.

$y (x + 2) (x - 2) = \pm e^{C} - 3 x^{2} + 12$

Step 3.10.7

Divide each term in $y (x + 2) (x - 2) = \pm e^{C} - 3 x^{2} + 12$ by $(x + 2) (x - 2)$ and simplify.

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Step 3.10.7.1

Divide each term in $y (x + 2) (x - 2) = \pm e^{C} - 3 x^{2} + 12$ by $(x + 2) (x - 2)$ .

$\frac{y (x + 2) (x - 2)}{(x + 2) (x - 2)} = \frac{\pm e^{C}}{(x + 2) (x - 2)} + \frac{- 3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$

Step 3.10.7.2

Simplify the left side.

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Step 3.10.7.2.1

Cancel the common factor of $x + 2$ .

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Step 3.10.7.2.1.1

Cancel the common factor.

$\frac{y (x + 2) (x - 2)}{(x + 2) (x - 2)} = \frac{\pm e^{C}}{(x + 2) (x - 2)} + \frac{- 3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$

Step 3.10.7.2.1.2

Rewrite the expression.

$\frac{y (x - 2)}{x - 2} = \frac{\pm e^{C}}{(x + 2) (x - 2)} + \frac{- 3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$

Step 3.10.7.2.2

Cancel the common factor of $x - 2$ .

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Step 3.10.7.2.2.1

Cancel the common factor.

$\frac{y (x - 2)}{x - 2} = \frac{\pm e^{C}}{(x + 2) (x - 2)} + \frac{- 3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$

Step 3.10.7.2.2.2

Divide $y$ by $1$ .

$y = \frac{\pm e^{C}}{(x + 2) (x - 2)} + \frac{- 3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$

Step 3.10.7.3

Simplify the right side.

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Step 3.10.7.3.1

Move the negative in front of the fraction.

$y = \frac{\pm e^{C}}{(x + 2) (x - 2)} - \frac{3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$

Step 4

Simplify the constant of integration.

$y = \frac{C}{(x + 2) (x - 2)} - \frac{3 x^{2}}{(x + 2) (x - 2)} + \frac{12}{(x + 2) (x - 2)}$