Calculus Examples

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

Step 1

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

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Simplify each term.

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Step 1.1.2.1

Cancel the common factor of $y^{2}$ and $y$ .

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

Cancel the common factors.

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Step 1.1.2.1.2.1

Factor $y$ out of $x y$ .

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

Step 1.1.2.1.2.2

Cancel the common factor.

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

Step 1.1.2.1.2.3

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $x$ .

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Step 1.1.2.2.1

Cancel the common factor.

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

Step 1.1.2.2.2

Rewrite the expression.

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

Step 1.2

Assume $\sqrt{y^{2}} = y$ .

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

Step 1.3

Combine $\sqrt{x^{2} + y^{2}}$ and $\sqrt{y^{2}}$ into a single radical.

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

Step 1.4

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

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Step 1.4.1

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

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

Step 1.4.2

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

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Step 1.4.2.1

Cancel the common factor.

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

Step 1.4.2.2

Rewrite the expression.

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

Step 1.5

Rewrite the differential equation as $f (\frac{y}{x})$ .

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Step 1.5.1

Rewrite $\frac{x^{2}}{y^{2}}$ as ${(\frac{x}{y})}^{2}$ .

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

Step 1.5.2

Rewrite $\frac{x}{y}$ as ${(\frac{y}{x})}^{- 1}$ .

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

Step 2

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

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

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

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

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

Simplify $V + \sqrt{{(V^{- 1})}^{2} + 1}$ .

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Step 6.1.1.1.1

Simplify each term.

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Step 6.1.1.1.1.1

Multiply the exponents in ${(V^{- 1})}^{2}$ .

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Step 6.1.1.1.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.1.1.1.1.1.2

Multiply $- 1$ by $2$ .

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

Step 6.1.1.1.1.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 6.1.1.1.1.3

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

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

Step 6.1.1.1.1.4

Combine the numerators over the common denominator.

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

Step 6.1.1.1.1.5

Rewrite $\frac{1 + V^{2}}{V^{2}}$ as ${(\frac{1}{V})}^{2} (1 + V^{2})$ .

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Step 6.1.1.1.1.5.1

Factor the perfect power $1^{2}$ out of $1 + V^{2}$ .

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

Step 6.1.1.1.1.5.2

Factor the perfect power $V^{2}$ out of $V^{2}$ .

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

Step 6.1.1.1.1.5.3

Rearrange the fraction $\frac{1^{2} (1 + V^{2})}{V^{2} \cdot 1}$ .

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

Step 6.1.1.1.1.6

Pull terms out from under the radical.

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

Step 6.1.1.1.1.7

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

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

Step 6.1.1.1.2

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

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

Step 6.1.1.1.3

Combine the numerators over the common denominator.

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

Step 6.1.1.1.4

Multiply $V$ by $V$ .

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

Step 6.1.1.2

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

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Step 6.1.1.2.1

Subtract $V$ from both sides of the equation.

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

Step 6.1.1.2.2

Simplify each term.

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Step 6.1.1.2.2.1

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

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

Step 6.1.1.2.2.2

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

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Step 6.1.1.2.2.2.1

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

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

Step 6.1.1.2.2.2.2

Cancel the common factors.

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Step 6.1.1.2.2.2.2.1

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

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

Step 6.1.1.2.2.2.2.2

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

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

Step 6.1.1.2.2.2.2.3

Cancel the common factor.

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

Step 6.1.1.2.2.2.2.4

Rewrite the expression.

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

Step 6.1.1.2.2.2.2.5

Divide $V$ by $1$ .

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

Step 6.1.1.2.3

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

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Step 6.1.1.2.3.1

Subtract $V$ from $V$ .

$x \frac{d V}{d x} = \frac{\sqrt{1 + V^{2}}}{V} + 0$

Step 6.1.1.2.3.2

Add $\frac{\sqrt{1 + V^{2}}}{V}$ and $0$ .

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

Step 6.1.1.3

Divide each term in $x \frac{d V}{d x} = \frac{\sqrt{1 + V^{2}}}{V}$ by $x$ and simplify.

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Step 6.1.1.3.1

Divide each term in $x \frac{d V}{d x} = \frac{\sqrt{1 + V^{2}}}{V}$ by $x$ .

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

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.

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

Step 6.1.1.3.2.1.2

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

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

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.1.1.3.3.2

Combine.

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

Step 6.1.1.3.3.3

Multiply $\sqrt{1 + V^{2}}$ by $1$ .

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

Step 6.1.2

Regroup factors.

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

Step 6.1.3

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

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

Step 6.1.4

Simplify.

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Step 6.1.4.1

Multiply $\frac{V}{\sqrt{1 + V^{2}}}$ by $\frac{\sqrt{1 + V^{2}}}{\sqrt{1 + V^{2}}}$ .

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

Step 6.1.4.2

Combine and simplify the denominator.

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Step 6.1.4.2.1

Multiply $\frac{V}{\sqrt{1 + V^{2}}}$ by $\frac{\sqrt{1 + V^{2}}}{\sqrt{1 + V^{2}}}$ .

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

Step 6.1.4.2.2

Raise $\sqrt{1 + V^{2}}$ to the power of $1$ .

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

Step 6.1.4.2.3

Raise $\sqrt{1 + V^{2}}$ to the power of $1$ .

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

Step 6.1.4.2.4

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

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

Step 6.1.4.2.5

Add $1$ and $1$ .

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

Step 6.1.4.2.6

Rewrite ${\sqrt{1 + V^{2}}}^{2}$ as $1 + V^{2}$ .

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Step 6.1.4.2.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + V^{2}}$ as ${(1 + V^{2})}^{\frac{1}{2}}$ .

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

Step 6.1.4.2.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.1.4.2.6.3

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

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

Step 6.1.4.2.6.4

Cancel the common factor of $2$ .

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Step 6.1.4.2.6.4.1

Cancel the common factor.

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

Step 6.1.4.2.6.4.2

Rewrite the expression.

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

Step 6.1.4.2.6.5

Simplify.

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

Step 6.1.4.3

Combine.

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

Step 6.1.4.4

Cancel the common factor of $V$ .

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Step 6.1.4.4.1

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

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

Step 6.1.4.4.2

Factor $V$ out of $V x$ .

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

Step 6.1.4.4.3

Cancel the common factor.

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

Step 6.1.4.4.4

Rewrite the expression.

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

Step 6.1.4.5

Multiply $\sqrt{1 + V^{2}}$ by $1$ .

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

Step 6.1.4.6

Multiply $\frac{\sqrt{1 + V^{2}}}{1 + V^{2}}$ by $\frac{\sqrt{1 + V^{2}}}{x}$ .

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

Step 6.1.4.7

Raise $\sqrt{1 + V^{2}}$ to the power of $1$ .

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

Step 6.1.4.8

Raise $\sqrt{1 + V^{2}}$ to the power of $1$ .

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

Step 6.1.4.9

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

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

Step 6.1.4.10

Add $1$ and $1$ .

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

Step 6.1.4.11

Rewrite ${\sqrt{1 + V^{2}}}^{2}$ as $1 + V^{2}$ .

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Step 6.1.4.11.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + V^{2}}$ as ${(1 + V^{2})}^{\frac{1}{2}}$ .

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

Step 6.1.4.11.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.1.4.11.3

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

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

Step 6.1.4.11.4

Cancel the common factor of $2$ .

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Step 6.1.4.11.4.1

Cancel the common factor.

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

Step 6.1.4.11.4.2

Rewrite the expression.

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

Step 6.1.4.11.5

Simplify.

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

Step 6.1.4.12

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

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Step 6.1.4.12.1

Cancel the common factor.

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

Step 6.1.4.12.2

Rewrite the expression.

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

Step 6.1.5

Rewrite the equation.

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

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

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

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

Let $u = 1 + V^{2}$ . 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 6.2.2.1.1

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

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Step 6.2.2.1.1.1

Differentiate $1 + V^{2}$ .

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

Step 6.2.2.1.1.2

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

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

Step 6.2.2.1.1.3

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

$0 + \frac{d}{d V} [V^{2}]$

Step 6.2.2.1.1.4

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

$0 + 2 V$

Step 6.2.2.1.1.5

Add $0$ and $2 V$ .

$2 V$

Step 6.2.2.1.2

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

$\int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} d u = \int \frac{1}{x} d x$

Step 6.2.2.2

Simplify.

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Step 6.2.2.2.1

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

$\int \frac{1}{\sqrt{u} \cdot 2} d u = \int \frac{1}{x} d x$

Step 6.2.2.2.2

Move $2$ to the left of $\sqrt{u}$ .

$\int \frac{1}{2 \sqrt{u}} d u = \int \frac{1}{x} d x$

Step 6.2.2.3

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

$\frac{1}{2} \int \frac{1}{\sqrt{u}} d u = \int \frac{1}{x} d x$

Step 6.2.2.4

Apply basic rules of exponents.

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Step 6.2.2.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u}$ as $u^{\frac{1}{2}}$ .

$\frac{1}{2} \int \frac{1}{u^{\frac{1}{2}}} d u = \int \frac{1}{x} d x$

Step 6.2.2.4.2

Move $u^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$\frac{1}{2} \int {(u^{\frac{1}{2}})}^{- 1} d u = \int \frac{1}{x} d x$

Step 6.2.2.4.3

Multiply the exponents in ${(u^{\frac{1}{2}})}^{- 1}$ .

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Step 6.2.2.4.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{2} \int u^{\frac{1}{2} \cdot - 1} d u = \int \frac{1}{x} d x$

Step 6.2.2.4.3.2

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

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

Step 6.2.2.4.3.3

Move the negative in front of the fraction.

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

Step 6.2.2.5

By the Power Rule, the integral of $u^{- \frac{1}{2}}$ with respect to $u$ is $2 u^{\frac{1}{2}}$ .

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

Step 6.2.2.6

Simplify.

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Step 6.2.2.6.1

Rewrite $\frac{1}{2} (2 u^{\frac{1}{2}} + C_{1})$ as $\frac{1}{2} \cdot 2 u^{\frac{1}{2}} + C_{1}$ .

$\frac{1}{2} \cdot 2 u^{\frac{1}{2}} + C_{1} = \int \frac{1}{x} d x$

Step 6.2.2.6.2

Simplify.

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Step 6.2.2.6.2.1

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

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

Step 6.2.2.6.2.2

Cancel the common factor of $2$ .

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Step 6.2.2.6.2.2.1

Cancel the common factor.

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

Step 6.2.2.6.2.2.2

Rewrite the expression.

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

Step 6.2.2.6.2.3

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

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

Step 6.2.2.7

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

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

Step 6.2.3

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

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

Step 6.2.4

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

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

Step 6.3

Solve for $V$ .

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Step 6.3.1

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

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

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

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

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Step 6.3.2.1.1

Multiply the exponents in ${({(1 + V^{2})}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 6.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.3.2.1.1.2.1

Cancel the common factor.

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

Step 6.3.2.1.1.2.2

Rewrite the expression.

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

Step 6.3.2.1.2

Simplify.

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

Step 6.3.3

Solve for $V$ .

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Step 6.3.3.1

Subtract $1$ from both sides of the equation.

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

Step 6.3.3.2

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

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

Step 6.3.3.3

Simplify $\pm \sqrt{{(\ln (| x |) + C)}^{2} - 1}$ .

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Step 6.3.3.3.1

Rewrite $1$ as $1^{2}$ .

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

Step 6.3.3.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \ln (| x |) + C$ and $b = 1$ .

$V = \pm \sqrt{(\ln (| x |) + C + 1) (\ln (| x |) + C - 1)}$

Step 6.3.3.4

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

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Step 6.3.3.4.1

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

$V = \sqrt{(\ln (| x |) + C + 1) (\ln (| x |) + C - 1)}$

Step 6.3.3.4.2

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