Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Apply the distributive property.

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

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

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

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

Step 1.4.2

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

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

Step 1.4.3

Cancel the common factor.

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

Step 1.4.4

Rewrite the expression.

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

Multiply $3 y^{2} \frac{1}{x^{2}}$ .

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Step 1.7.1

Combine $3$ and $\frac{1}{x^{2}}$ .

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

Step 1.7.2

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

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

Step 1.8

Apply the distributive property.

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

Step 1.9

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

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Step 1.9.1

Cancel the common factor.

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

Step 1.9.2

Rewrite the expression.

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

Step 1.10

Cancel the common factor of $x$ .

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Step 1.10.1

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

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

Step 1.10.2

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

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

Step 1.10.3

Cancel the common factor.

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

Step 1.10.4

Rewrite the expression.

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

Step 1.11

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

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

Step 1.12

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

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

Step 1.13

Simplify each term.

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Step 1.13.1

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

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

Step 1.13.2

Move $3$ to the left of $y^{2}$ .

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

Step 1.14

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

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

Step 1.15

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

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Step 1.15.1

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

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

Step 1.15.2

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

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

Step 1.16

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

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Step 1.16.1

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

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

Step 1.16.2

Reorder $\frac{y}{x}$ and $\frac{3 y}{x}$ .

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

Step 1.17

Factor out $\frac{y}{x}$ from $\frac{3 y}{x}$ .

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Step 1.17.1

Factor $\frac{y}{x}$ out of $\frac{3 y}{x}$ .

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

Step 1.17.2

Reorder $\frac{y}{x}$ and $3$ .

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

Step 1.18

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

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Step 1.18.1

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

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

Step 1.18.2

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

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

Step 2

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

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

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

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

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

Factor $V$ out of $2 \cdot V + 3 \cdot V \cdot V$ .

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Step 6.1.1.1.1

Factor $V$ out of $2 \cdot V$ .

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

Step 6.1.1.1.2

Factor $V$ out of $3 \cdot V \cdot V$ .

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

Step 6.1.1.1.3

Factor $V$ out of $V \cdot 2 + V (3 V)$ .

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

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

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{V (2 + 3 V)}{1 + 2 V} - V$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{V (2 + 3 V)}{1 + 2 V} - V$ by $x$ .

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

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.

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

Step 6.1.1.3.2.1.2

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

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

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.

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

Step 6.1.1.3.3.2

To write $- V$ as a fraction with a common denominator, multiply by $\frac{1 + 2 V}{1 + 2 V}$ .

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

Step 6.1.1.3.3.3

Simplify terms.

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Step 6.1.1.3.3.3.1

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

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

Step 6.1.1.3.3.3.2

Combine the numerators over the common denominator.

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

Step 6.1.1.3.3.4

Simplify the numerator.

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Step 6.1.1.3.3.4.1

Factor $V$ out of $V (2 + 3 V) - V (1 + 2 V)$ .

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Step 6.1.1.3.3.4.1.1

Factor $V$ out of $- V (1 + 2 V)$ .

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

Step 6.1.1.3.3.4.1.2

Factor $V$ out of $V (2 + 3 V) + V (- (1 + 2 V))$ .

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

Step 6.1.1.3.3.4.2

Apply the distributive property.

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

Step 6.1.1.3.3.4.3

Multiply $- 1$ by $1$ .

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

Step 6.1.1.3.3.4.4

Multiply $2$ by $- 1$ .

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

Step 6.1.1.3.3.4.5

Subtract $1$ from $2$ .

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

Step 6.1.1.3.3.4.6

Subtract $2 V$ from $3 V$ .

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

Step 6.1.1.3.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.6

Multiply $\frac{V (V + 1)}{1 + 2 V}$ by $\frac{1}{x}$ .

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

Step 6.1.1.3.3.7

Reorder factors in $\frac{V (V + 1)}{(1 + 2 V) x}$ .

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

Step 6.1.2

Regroup factors.

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

Step 6.1.3

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

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

Multiply $\frac{V (V + 1)}{1 + 2 V}$ by $\frac{1}{x}$ .

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

Step 6.1.4.2

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

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Step 6.1.4.2.1

Factor $1 + 2 V$ out of $(1 + 2 V) x$ .

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

Step 6.1.4.2.2

Cancel the common factor.

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

Step 6.1.4.2.3

Rewrite the expression.

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

Step 6.1.4.3

Cancel the common factor of $V (V + 1)$ .

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Step 6.1.4.3.1

Cancel the common factor.

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

Step 6.1.4.3.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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 in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{V} + \frac{B}{V + 1}$

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 $V (V + 1)$ .

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

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.

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

Step 6.2.2.1.1.3.2

Rewrite the expression.

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

Step 6.2.2.1.1.4

Cancel the common factor of $V + 1$ .

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Step 6.2.2.1.1.4.1

Cancel the common factor.

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

Step 6.2.2.1.1.4.2

Divide $1 + 2 V$ by $1$ .

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

Step 6.2.2.1.1.5

Reorder $1$ and $2 V$ .

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

Step 6.2.2.1.1.6

Simplify each term.

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Step 6.2.2.1.1.6.1

Cancel the common factor of $V$ .

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Step 6.2.2.1.1.6.1.1

Cancel the common factor.

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

Step 6.2.2.1.1.6.1.2

Divide $A (V + 1)$ by $1$ .

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

Step 6.2.2.1.1.6.2

Apply the distributive property.

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

Step 6.2.2.1.1.6.3

Multiply $A$ by $1$ .

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

Step 6.2.2.1.1.6.4

Cancel the common factor of $V + 1$ .

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Step 6.2.2.1.1.6.4.1

Cancel the common factor.

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

Step 6.2.2.1.1.6.4.2

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

$2 V + 1 = A V + A + (B) (V)$

$2 V + 1 = A V + A + B V$

Step 6.2.2.1.1.7

Move $A$ .

$2 V + 1 = A V + B V + A$

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 $V$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$2 = A + B$

Step 6.2.2.1.2.2

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 6.2.2.1.2.3

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

$2 = A + B$

$1 = A$

$2 = A + B$

$1 = A$

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 $A = 1$ .

$A = 1$

$2 = A + B$

Step 6.2.2.1.3.2

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

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Step 6.2.2.1.3.2.1

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

$2 = (1) + B$

$A = 1$

Step 6.2.2.1.3.2.2

Simplify the right side.

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Step 6.2.2.1.3.2.2.1

Remove parentheses.

$2 = 1 + B$

$A = 1$

$2 = 1 + B$

$A = 1$

$2 = 1 + B$

$A = 1$

Step 6.2.2.1.3.3

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

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Step 6.2.2.1.3.3.1

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

$1 + B = 2$

$A = 1$

Step 6.2.2.1.3.3.2

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

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Step 6.2.2.1.3.3.2.1

Subtract $1$ from both sides of the equation.

$B = 2 - 1$

$A = 1$

Step 6.2.2.1.3.3.2.2

Subtract $1$ from $2$ .

$B = 1$

$A = 1$

$B = 1$

$A = 1$

$B = 1$

$A = 1$

Step 6.2.2.1.3.4

Solve the system of equations.

$B = 1 A = 1$

Step 6.2.2.1.3.5

List all of the solutions.

$B = 1, A = 1$

Step 6.2.2.1.4

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

$\frac{1}{V} + \frac{1}{V + 1}$

Step 6.2.2.1.5

Remove the zero from the expression.

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

Step 6.2.2.2

Split the single integral into multiple integrals.

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

Step 6.2.2.3

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

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

Step 6.2.2.4

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

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Step 6.2.2.4.1

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

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Step 6.2.2.4.1.1

Differentiate $V + 1$ .

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

Step 6.2.2.4.1.2

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 6.2.2.4.1.3

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 6.2.2.4.1.4

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

$1 + 0$

Step 6.2.2.4.1.5

Add $1$ and $0$ .

$1$

Step 6.2.2.4.2

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

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

Step 6.2.2.5

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

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

Step 6.2.2.6

Simplify.

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Step 6.2.2.6.1

Simplify.

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

Step 6.2.2.6.2

Simplify.

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Step 6.2.2.6.2.1

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

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

Step 6.2.2.6.2.2

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

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

Step 6.2.2.7

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

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

Simplify the numerator.

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Step 8.3.1

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

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

Step 8.3.2

Simplify the numerator.

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Step 8.3.2.1

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

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

Step 8.3.2.2

Combine the numerators over the common denominator.

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

Step 8.3.3

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

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

Step 8.3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.3.5

Multiply $\frac{y (y + x)}{x} \cdot \frac{1}{x}$ .

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Step 8.3.5.1

Multiply $\frac{y (y + x)}{x}$ by $\frac{1}{x}$ .

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

Step 8.3.5.2

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

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

Step 8.3.5.3

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

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

Step 8.3.5.4

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

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

Step 8.3.5.5

Add $1$ and $1$ .

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

Step 8.3.6

Remove non-negative terms from the absolute value.

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

Step 8.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.5

Combine.

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

Step 8.6

Multiply $| y (y + x) |$ by $1$ .

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