Calculus Examples

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

Step 1

Rewrite the differential equation as a function of $\frac{y}{x}$ .

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Apply the distributive property.

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

Step 1.4

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

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

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

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

Step 1.6

Apply the distributive property.

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

Step 1.7

Cancel the common factor of $x$ .

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Step 1.7.1

Factor $x$ out of $x y$ .

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

Step 1.7.2

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

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

Step 1.7.3

Cancel the common factor.

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

Step 1.7.4

Rewrite the expression.

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

Step 1.8

Combine $y$ and $\frac{1}{x}$ .

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

Step 1.9

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

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Step 1.9.1

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

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

Step 1.9.2

Cancel the common factor.

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

Step 1.9.3

Rewrite the expression.

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

Step 1.10

Use the power of quotient rule $\frac{a^{m}}{b^{m}} = {(\frac{a}{b})}^{m}$ .

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

Step 2

Let $V = \frac{y}{x}$ . Substitute $V$ for $\frac{y}{x}$ .

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

Step 3

Solve $V = \frac{y}{x}$ for $y$ .

$y = V x$

Step 4

Use the product rule to find the derivative of $y = V x$ with respect to $x$ .

$\frac{d y}{d x} = x \frac{d V}{d x} + V$

Step 5

Substitute $x \frac{d V}{d x} + V$ for $\frac{d y}{d x}$ .

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

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

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

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Step 6.1.1.1

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

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Step 6.1.1.1.1

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.1.2

Split the fraction $\frac{1 + V^{2}}{V - 1}$ into two fractions.

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

Step 6.1.1.1.3

Find the common denominator.

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Step 6.1.1.1.3.1

Write $- V$ as a fraction with denominator $1$ .

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

Step 6.1.1.1.3.2

Multiply $\frac{- V}{1}$ by $\frac{V - 1}{V - 1}$ .

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

Step 6.1.1.1.3.3

Multiply $\frac{- V}{1}$ by $\frac{V - 1}{V - 1}$ .

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

Step 6.1.1.1.4

Combine the numerators over the common denominator.

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

Step 6.1.1.1.5

Simplify each term.

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Step 6.1.1.1.5.1

Apply the distributive property.

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

Step 6.1.1.1.5.2

Multiply $V$ by $V$ by adding the exponents.

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Step 6.1.1.1.5.2.1

Move $V$ .

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

Step 6.1.1.1.5.2.2

Multiply $V$ by $V$ .

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

Step 6.1.1.1.5.3

Multiply $- V \cdot - 1$ .

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Step 6.1.1.1.5.3.1

Multiply $- 1$ by $- 1$ .

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

Step 6.1.1.1.5.3.2

Multiply $V$ by $1$ .

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

Step 6.1.1.1.6

Combine the opposite terms in $1 + V^{2} - V^{2} + V$ .

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Step 6.1.1.1.6.1

Subtract $V^{2}$ from $V^{2}$ .

$x \frac{d V}{d x} = \frac{1 + 0 + V}{V - 1}$

Step 6.1.1.1.6.2

Add $1$ and $0$ .

$x \frac{d V}{d x} = \frac{1 + V}{V - 1}$

Step 6.1.1.1.7

Split the fraction $\frac{1 + V}{V - 1}$ into two fractions.

$x \frac{d V}{d x} = \frac{1}{V - 1} + \frac{V}{V - 1}$

Step 6.1.1.2

Divide each term in $x \frac{d V}{d x} = \frac{1}{V - 1} + \frac{V}{V - 1}$ by $x$ and simplify.

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Step 6.1.1.2.1

Divide each term in $x \frac{d V}{d x} = \frac{1}{V - 1} + \frac{V}{V - 1}$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{1}{V - 1}}{x} + \frac{\frac{V}{V - 1}}{x}$

Step 6.1.1.2.2

Simplify the left side.

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Step 6.1.1.2.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.2.2.1.1

Cancel the common factor.

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{1}{V - 1}}{x} + \frac{\frac{V}{V - 1}}{x}$

Step 6.1.1.2.2.1.2

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

$\frac{d V}{d x} = \frac{\frac{1}{V - 1}}{x} + \frac{\frac{V}{V - 1}}{x}$

Step 6.1.1.2.3

Simplify the right side.

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Step 6.1.1.2.3.1

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{1}{V - 1} + \frac{V}{V - 1}}{x}$

Step 6.1.1.2.3.2

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{1 + V}{V - 1}}{x}$

Step 6.1.1.2.3.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{d V}{d x} = \frac{1 + V}{V - 1} \cdot \frac{1}{x}$

Step 6.1.1.2.3.4

Multiply $\frac{1 + V}{V - 1}$ by $\frac{1}{x}$ .

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

Step 6.1.1.2.3.5

Reorder factors in $\frac{1 + V}{(V - 1) x}$ .

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

Step 6.1.2

Regroup factors.

$\frac{d V}{d x} = \frac{1 + V}{V - 1} \cdot \frac{1}{x}$

Step 6.1.3

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

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

Multiply $\frac{1 + V}{V - 1}$ by $\frac{1}{x}$ .

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

Step 6.1.4.2

Cancel the common factor of $V - 1$ .

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Step 6.1.4.2.1

Factor $V - 1$ out of $(V - 1) x$ .

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

Step 6.1.4.2.2

Cancel the common factor.

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

Step 6.1.4.2.3

Rewrite the expression.

$\frac{V - 1}{1 + V} \frac{d V}{d x} = \frac{1}{1 + V} \cdot \frac{1 + V}{x}$

Step 6.1.4.3

Cancel the common factor of $1 + V$ .

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Step 6.1.4.3.1

Cancel the common factor.

$\frac{V - 1}{1 + V} \frac{d V}{d x} = \frac{1}{1 + V} \cdot \frac{1 + V}{x}$

Step 6.1.4.3.2

Rewrite the expression.

$\frac{V - 1}{1 + V} \frac{d V}{d x} = \frac{1}{x}$

Step 6.1.5

Rewrite the equation.

$\frac{V - 1}{1 + V} d V = \frac{1}{x} d x$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{V - 1}{1 + V} d V = \int \frac{1}{x} d x$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Reorder $1$ and $V$ .

$\int \frac{V - 1}{V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.2

Divide $V - 1$ by $V + 1$ .

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Step 6.2.2.2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$V$

$1$

$V$

$1$

Step 6.2.2.2.2

Divide the highest order term in the dividend $V$ by the highest order term in divisor $V$ .

					$1$
	$V$	+	$1$		$V$	-	$1$

Step 6.2.2.2.3

Multiply the new quotient term by the divisor.

				$1$
$V$	+	$1$		$V$	-	$1$
			+	$V$	+	$1$

Step 6.2.2.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $V + 1$

				$1$
$V$	+	$1$		$V$	-	$1$
			-	$V$	-	$1$

Step 6.2.2.2.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$V$	+	$1$		$V$	-	$1$
			-	$V$	-	$1$
					-	$2$

Step 6.2.2.2.6

The final answer is the quotient plus the remainder over the divisor.

$\int 1 - \frac{2}{V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.3

Split the single integral into multiple integrals.

$\int d V + \int - \frac{2}{V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.4

Apply the constant rule.

$V + C_{1} + \int - \frac{2}{V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.5

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

$V + C_{1} - \int \frac{2}{V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.6

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

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

Step 6.2.2.7

Multiply $2$ by $- 1$ .

$V + C_{1} - 2 \int \frac{1}{V + 1} d V = \int \frac{1}{x} d x$

Step 6.2.2.8

Let $u = V + 1$ . Then $d u = d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.8.1

Let $u = V + 1$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.8.1.1

Differentiate $V + 1$ .

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

Step 6.2.2.8.1.2

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

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

Step 6.2.2.8.1.3

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

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

Step 6.2.2.8.1.4

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

$1 + 0$

Step 6.2.2.8.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.8.2

Rewrite the problem using $u$ and $d u$ .

$V + C_{1} - 2 \int \frac{1}{u} d u = \int \frac{1}{x} d x$

Step 6.2.2.9

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

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

Step 6.2.2.10

Simplify.

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

Step 6.2.2.11

Replace all occurrences of $u$ with $V + 1$ .

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

Step 6.2.3

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

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

Step 6.2.4

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

$V - 2 \ln (| V + 1 |) = \ln (| x |) + C$

Step 7

Substitute $\frac{y}{x}$ for $V$ .

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

Step 8

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

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Step 8.1

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

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

Step 8.2

Simplify the left side.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

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

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

Step 8.2.1.2

Remove the absolute value in ${| \frac{y}{x} + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 8.3

Reorder $- \frac{y}{x}$ and $C$ .

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

Step 8.4

Move all terms containing $y$ to the left side of the equation.

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Step 8.4.1

Add $\frac{y}{x}$ to both sides of the equation.

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

Step 8.4.2

To write $- \ln ({(\frac{y}{x} + 1)}^{2})$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 8.4.3

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

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

Step 8.4.4

Combine the numerators over the common denominator.

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

Step 8.4.5

Simplify each term.

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Step 8.4.5.1

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

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

Step 8.4.5.2

Combine the numerators over the common denominator.

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

Step 8.4.5.3

Apply the product rule to $\frac{y + x}{x}$ .

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

Step 8.4.6

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

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

Step 8.4.7

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

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

Step 8.4.8

Combine the numerators over the common denominator.

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

Step 8.4.9

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

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

Step 8.4.10

Factor $- 1$ out of $y$ .

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

Step 8.4.11

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

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

Step 8.4.12

Factor $- 1$ out of $- \ln (| x |) x$ .

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

Step 8.4.13

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

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

Step 8.4.14

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

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

Step 8.4.15

Move the negative in front of the fraction.

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

Step 8.4.16

Reorder factors in $- \frac{\ln (\frac{{(y + x)}^{2}}{x^{2}}) x - y + \ln (| x |) x}{x}$ .

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