Calculus Examples

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

Step 1

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

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Step 1.1

Divide each term in $2 x^{3} \frac{d y}{d x} = y (y^{2} + 3 x^{2})$ by $2 x^{3}$ and simplify.

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Step 1.1.1

Divide each term in $2 x^{3} \frac{d y}{d x} = y (y^{2} + 3 x^{2})$ by $2 x^{3}$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $2$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $x^{3}$ .

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Step 1.1.2.2.1

Cancel the common factor.

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

Step 1.1.2.2.2

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

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

Step 1.2

Factor out $\frac{y}{x}$ from $\frac{y (y^{2} + 3 x^{2})}{2 x^{3}}$ .

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Step 1.2.1

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

$\frac{d y}{d x} = \frac{y}{x} \cdot \frac{y^{2} + 3 x^{2}}{2 x^{2}}$

Step 1.2.2

Reorder $\frac{y}{x}$ and $\frac{y^{2} + 3 x^{2}}{2 x^{2}}$ .

$\frac{d y}{d x} = \frac{y^{2} + 3 x^{2}}{2 x^{2}} \cdot \frac{y}{x}$

Step 1.3

Split $\frac{y^{2} + 3 x^{2}}{2 x^{2}}$ and simplify.

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Step 1.3.1

Split the fraction $\frac{y^{2} + 3 x^{2}}{2 x^{2}}$ into two fractions.

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

Step 1.3.2

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

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Step 1.3.2.1

Cancel the common factor.

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

Step 1.3.2.2

Rewrite the expression.

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

Reorder $\frac{y}{x}$ and $\frac{y}{2 x}$ .

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

Step 1.5

Factor out $\frac{y}{x}$ from $\frac{y}{2 x}$ .

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Step 1.5.1

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

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

Step 1.5.2

Reorder $\frac{y}{x}$ and $\frac{1}{2}$ .

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

Step 2

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

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

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

$x \frac{d V}{d x} + V = (\frac{1}{2} V \cdot V + \frac{3}{2}) V$

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

Simplify $(\frac{1}{2} \cdot (V \cdot V) + \frac{3}{2}) V$ .

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Step 6.1.1.1.1

Rewrite.

$x \frac{d V}{d x} + V = 0 + 0 + (\frac{1}{2} \cdot (V \cdot V) + \frac{3}{2}) V$

Step 6.1.1.1.2

Simplify by adding zeros.

$x \frac{d V}{d x} + V = (\frac{1}{2} \cdot (V \cdot V) + \frac{3}{2}) V$

Step 6.1.1.1.3

Simplify each term.

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Step 6.1.1.1.3.1

Multiply $V$ by $V$ .

$x \frac{d V}{d x} + V = (\frac{1}{2} \cdot V^{2} + \frac{3}{2}) V$

Step 6.1.1.1.3.2

Combine $\frac{1}{2}$ and $V^{2}$ .

$x \frac{d V}{d x} + V = (\frac{V^{2}}{2} + \frac{3}{2}) V$

Step 6.1.1.1.4

Apply the distributive property.

$x \frac{d V}{d x} + V = \frac{V^{2}}{2} V + \frac{3}{2} V$

Step 6.1.1.1.5

Multiply $\frac{V^{2}}{2} V$ .

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Step 6.1.1.1.5.1

Combine $\frac{V^{2}}{2}$ and $V$ .

$x \frac{d V}{d x} + V = \frac{V^{2} V}{2} + \frac{3}{2} V$

Step 6.1.1.1.5.2

Multiply $V^{2}$ by $V$ by adding the exponents.

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Step 6.1.1.1.5.2.1

Multiply $V^{2}$ by $V$ .

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Step 6.1.1.1.5.2.1.1

Raise $V$ to the power of $1$ .

$x \frac{d V}{d x} + V = \frac{V^{2} V^{1}}{2} + \frac{3}{2} V$

Step 6.1.1.1.5.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x \frac{d V}{d x} + V = \frac{V^{2 + 1}}{2} + \frac{3}{2} V$

Step 6.1.1.1.5.2.2

Add $2$ and $1$ .

$x \frac{d V}{d x} + V = \frac{V^{3}}{2} + \frac{3}{2} V$

Step 6.1.1.1.6

Combine $\frac{3}{2}$ and $V$ .

$x \frac{d V}{d x} + V = \frac{V^{3}}{2} + \frac{3 V}{2}$

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

$x \frac{d V}{d x} = \frac{V^{3}}{2} + \frac{3 V}{2} - V$

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{V^{3}}{2} + \frac{3 V}{2} - V$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{V^{3}}{2} + \frac{3 V}{2} - V$ by $x$ .

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{V^{3}}{2}}{x} + \frac{\frac{3 V}{2}}{x} + \frac{- V}{x}$

Step 6.1.1.3.2

Simplify the left side.

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Step 6.1.1.3.2.1

Cancel the common factor of $x$ .

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Step 6.1.1.3.2.1.1

Cancel the common factor.

$\frac{x \frac{d V}{d x}}{x} = \frac{\frac{V^{3}}{2}}{x} + \frac{\frac{3 V}{2}}{x} + \frac{- V}{x}$

Step 6.1.1.3.2.1.2

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

$\frac{d V}{d x} = \frac{\frac{V^{3}}{2}}{x} + \frac{\frac{3 V}{2}}{x} + \frac{- V}{x}$

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{V^{3}}{2} + \frac{3 V}{2} - V}{x}$

Step 6.1.1.3.3.2

To write $- V$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{d V}{d x} = \frac{\frac{V^{3}}{2} + \frac{3 V}{2} - V \cdot \frac{2}{2}}{x}$

Step 6.1.1.3.3.3

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

$\frac{d V}{d x} = \frac{\frac{V^{3}}{2} + \frac{3 V}{2} + \frac{- V \cdot 2}{2}}{x}$

Step 6.1.1.3.3.4

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{V^{3}}{2} + \frac{3 V - V \cdot 2}{2}}{x}$

Step 6.1.1.3.3.5

Combine the numerators over the common denominator.

$\frac{d V}{d x} = \frac{\frac{V^{3} + 3 V - V \cdot 2}{2}}{x}$

Step 6.1.1.3.3.6

Multiply $2$ by $- 1$ .

$\frac{d V}{d x} = \frac{\frac{V^{3} + 3 V - 2 V}{2}}{x}$

Step 6.1.1.3.3.7

Subtract $2 V$ from $3 V$ .

$\frac{d V}{d x} = \frac{\frac{V^{3} + V}{2}}{x}$

Step 6.1.1.3.3.8

Factor $V$ out of $V^{3} + V$ .

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Step 6.1.1.3.3.8.1

Factor $V$ out of $V^{3}$ .

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

Step 6.1.1.3.3.8.2

Raise $V$ to the power of $1$ .

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

Step 6.1.1.3.3.8.3

Factor $V$ out of $V^{1}$ .

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

Step 6.1.1.3.3.8.4

Factor $V$ out of $V \cdot V^{2} + V \cdot 1$ .

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

Step 6.1.1.3.3.9

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.10

Combine.

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

Step 6.1.1.3.3.11

Multiply $V$ by $1$ .

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

Step 6.1.2

Multiply both sides by $\frac{1}{V (V^{2} + 1)}$ .

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

Step 6.1.3

Simplify.

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Step 6.1.3.1

Combine.

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

Step 6.1.3.2

Cancel the common factor of $V$ .

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Step 6.1.3.2.1

Cancel the common factor.

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

Step 6.1.3.2.2

Rewrite the expression.

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

Step 6.1.3.3

Cancel the common factor of $V^{2} + 1$ .

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Step 6.1.3.3.1

Cancel the common factor.

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

Step 6.1.3.3.2

Rewrite the expression.

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

Step 6.1.4

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Write the fraction using partial fraction decomposition.

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Step 6.2.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 6.2.2.1.1.1

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

$\frac{A}{V} + \frac{B V + C}{V^{2} + 1}$

Step 6.2.2.1.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $V (V^{2} + 1)$ .

$\frac{1 (V (V^{2} + 1))}{V (V^{2} + 1)} = \frac{A (V (V^{2} + 1))}{V} + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.3

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.3.1

Cancel the common factor.

$\frac{1 (V (V^{2} + 1))}{V (V^{2} + 1)} = \frac{A (V (V^{2} + 1))}{V} + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.3.2

Rewrite the expression.

$\frac{1 (V^{2} + 1)}{V^{2} + 1} = \frac{A (V (V^{2} + 1))}{V} + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.4

Cancel the common factor of $V^{2} + 1$ .

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Step 6.2.2.1.1.4.1

Cancel the common factor.

$\frac{1 (V^{2} + 1)}{V^{2} + 1} = \frac{A (V (V^{2} + 1))}{V} + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.4.2

Rewrite the expression.

$1 = \frac{A (V (V^{2} + 1))}{V} + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.5

Simplify each term.

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Step 6.2.2.1.1.5.1

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.5.1.1

Cancel the common factor.

$1 = \frac{A (V (V^{2} + 1))}{V} + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.5.1.2

Divide $A (V^{2} + 1)$ by $1$ .

$1 = A (V^{2} + 1) + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.5.2

Apply the distributive property.

$1 = A V^{2} + A \cdot 1 + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.5.3

Multiply $A$ by $1$ .

$1 = A V^{2} + A + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.5.4

Cancel the common factor of $V^{2} + 1$ .

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Step 6.2.2.1.1.5.4.1

Cancel the common factor.

$1 = A V^{2} + A + \frac{(B V + C) (V (V^{2} + 1))}{V^{2} + 1}$

Step 6.2.2.1.1.5.4.2

Divide $(B V + C) (V)$ by $1$ .

$1 = A V^{2} + A + (B V + C) (V)$

Step 6.2.2.1.1.5.5

Apply the distributive property.

$1 = A V^{2} + A + B V \cdot V + C V$

Step 6.2.2.1.1.5.6

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

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Step 6.2.2.1.1.5.6.1

Move $V$ .

$1 = A V^{2} + A + B (V \cdot V) + C V$

Step 6.2.2.1.1.5.6.2

Multiply $V$ by $V$ .

$1 = A V^{2} + A + B V^{2} + C V$

Step 6.2.2.1.1.6

Move $A$ .

$1 = A V^{2} + B V^{2} + C V + A$

Step 6.2.2.1.2

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

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Step 6.2.2.1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $V^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 6.2.2.1.2.2

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

$0 = C$

Step 6.2.2.1.2.3

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

$1 = A$

Step 6.2.2.1.2.4

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

$0 = A + B$

$0 = C$

$1 = A$

$0 = A + B$

$0 = C$

$1 = A$

Step 6.2.2.1.3

Solve the system of equations.

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Step 6.2.2.1.3.1

Rewrite the equation as $C = 0$ .

$C = 0$

$0 = A + B$

$1 = A$

Step 6.2.2.1.3.2

Rewrite the equation as $A = 1$ .

$A = 1$

$C = 0$

$0 = A + B$

Step 6.2.2.1.3.3

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

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Step 6.2.2.1.3.3.1

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

$0 = (1) + B$

$A = 1$

$C = 0$

Step 6.2.2.1.3.3.2

Simplify the right side.

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Step 6.2.2.1.3.3.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$C = 0$

$0 = 1 + B$

$A = 1$

$C = 0$

$0 = 1 + B$

$A = 1$

$C = 0$

Step 6.2.2.1.3.4

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

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Step 6.2.2.1.3.4.1

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

$1 + B = 0$

$A = 1$

$C = 0$

Step 6.2.2.1.3.4.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$C = 0$

$B = - 1$

$A = 1$

$C = 0$

Step 6.2.2.1.3.5

Solve the system of equations.

$B = - 1 A = 1 C = 0$

Step 6.2.2.1.3.6

List all of the solutions.

$B = - 1, A = 1, C = 0$

Step 6.2.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{V} + \frac{B V + C}{V^{2} + 1}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{1}{V} + \frac{- 1 V + 0}{V^{2} + 1}$

Step 6.2.2.1.5

Simplify.

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Step 6.2.2.1.5.1

Remove parentheses.

$\frac{1}{V} + \frac{(- 1) \cdot V + 0}{V^{2} + 1}$

Step 6.2.2.1.5.2

Simplify the numerator.

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Step 6.2.2.1.5.2.1

Rewrite $- 1 V$ as $- V$ .

$\frac{1}{V} + \frac{- V + 0}{V^{2} + 1}$

Step 6.2.2.1.5.2.2

Add $- V$ and $0$ .

$\frac{1}{V} + \frac{- V}{V^{2} + 1}$

Step 6.2.2.1.5.3

Move the negative in front of the fraction.

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

Step 6.2.2.2

Split the single integral into multiple integrals.

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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

Step 6.2.2.5

Let $u = V^{2} + 1$ . Then $d u = 2 V d V$ , so $\frac{1}{2} d u = V d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.5.1

Let $u = V^{2} + 1$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.5.1.1

Differentiate $V^{2} + 1$ .

$\frac{d}{d V} [V^{2} + 1]$

Step 6.2.2.5.1.2

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

$\frac{d}{d V} [V^{2}] + \frac{d}{d V} [1]$

Step 6.2.2.5.1.3

Differentiate using the Power Rule which states that $\frac{d}{d V} [V^{n}]$ is $n V^{n - 1}$ where $n = 2$ .

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

Step 6.2.2.5.1.4

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

$2 V + 0$

Step 6.2.2.5.1.5

Add $2 V$ and $0$ .

$2 V$

Step 6.2.2.5.2

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

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

Step 6.2.2.6

Simplify.

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Step 6.2.2.6.1

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

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

Step 6.2.2.6.2

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

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

Step 6.2.2.7

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

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

Step 6.2.2.8

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

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

Step 6.2.2.9

Simplify.

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

Step 6.2.2.10

Replace all occurrences of $u$ with $V^{2} + 1$ .

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

Step 6.2.2.11

Reorder terms.

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

Step 6.2.3

Integrate the right side.

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Step 6.2.3.1

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

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

Step 6.2.3.2

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

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

Step 6.2.3.3

Simplify.

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

Step 6.2.4

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

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

Step 7

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

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

Step 8

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

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Step 8.1

Simplify the expressions in the equation.

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Step 8.1.1

Simplify the left side.

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Step 8.1.1.1

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

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

Step 8.1.2

Simplify the right side.

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Step 8.1.2.1

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

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

Step 8.2

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

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Step 8.2.1

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

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

Step 8.2.2

Simplify the left side.

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Step 8.2.2.1

Simplify each term.

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Step 8.2.2.1.1

Cancel the common factor of $2$ .

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Step 8.2.2.1.1.1

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

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

Step 8.2.2.1.1.2

Cancel the common factor.

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

Step 8.2.2.1.1.3

Rewrite the expression.

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

Step 8.2.2.1.2

Move $2$ to the left of $\ln (| \frac{y}{x} |)$ .

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

Step 8.2.3

Simplify the right side.

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Step 8.2.3.1

Simplify each term.

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Step 8.2.3.1.1

Cancel the common factor of $2$ .

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Step 8.2.3.1.1.1

Cancel the common factor.

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

Step 8.2.3.1.1.2

Rewrite the expression.

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

Step 8.2.3.1.2

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

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

Simplify the left side.

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Step 8.5.1

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

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Step 8.5.1.1

Simplify each term.

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Step 8.5.1.1.1

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

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

Step 8.5.1.1.2

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

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

Step 8.5.1.1.3

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

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

Step 8.5.1.2

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

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

Step 8.5.1.3

Simplify each term.

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Step 8.5.1.3.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.5.1.3.2

Combine.

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

Step 8.5.1.3.3

Multiply $y^{2}$ by $1$ .

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

Step 8.6

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

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

Step 8.7

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

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

Step 8.8

Solve for $y$ .

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Step 8.8.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 8.8.2

Expand the left side.

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Step 8.8.2.1

Expand $\ln (e^{2 C + \ln (| \frac{y^{2}}{x^{2}} + 1 |)})$ by moving $2 C + \ln (| \frac{y^{2}}{x^{2}} + 1 |)$ outside the logarithm.

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

Step 8.8.2.2

The natural logarithm of $e$ is $1$ .

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

Step 8.8.2.3

Multiply $2 C + \ln (| \frac{y^{2}}{x^{2}} + 1 |)$ by $1$ .

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

Step 8.8.3

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

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

Step 8.8.4

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

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

Step 8.8.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.8.6

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

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Step 8.8.6.1

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

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

Step 8.8.6.2

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

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

Step 8.8.7

Simplify the numerator.

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Step 8.8.7.1

Apply the distributive property.

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

Step 8.8.7.2

Cancel the common factor of $x$ .

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Step 8.8.7.2.1

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

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

Step 8.8.7.2.2

Cancel the common factor.

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

Step 8.8.7.2.3

Rewrite the expression.

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

Step 8.8.7.3

Multiply $x$ by $1$ .

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

Step 8.8.8

Expand the left side.

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Step 8.8.8.1

Expand $\ln (e^{2 C + \ln (| \frac{y^{2}}{x^{2}} + 1 |)})$ by moving $2 C + \ln (| \frac{y^{2}}{x^{2}} + 1 |)$ outside the logarithm.

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

Step 8.8.8.2

The natural logarithm of $e$ is $1$ .

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

Step 8.8.8.3

Multiply $2 C + \ln (| \frac{y^{2}}{x^{2}} + 1 |)$ by $1$ .

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