Calculus Examples

$2 x v d v + (v^{2} - 1) d x = 0$

Step 1

Subtract $(v^{2} - 1) d x$ from both sides of the equation.

$2 x v d v = - (v^{2} - 1) d x$

Step 2

Multiply both sides by $\frac{1}{x (v^{2} - 1)}$ .

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Simplify the denominator.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.3

Combine $2$ and $\frac{1}{x (v + 1) (v - 1)}$ .

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

Step 3.4

Cancel the common factor of $x$ .

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Step 3.4.1

Factor $x$ out of $x (v + 1) (v - 1)$ .

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

Step 3.4.2

Factor $x$ out of $x v$ .

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

Step 3.4.3

Cancel the common factor.

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

Step 3.4.4

Rewrite the expression.

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

Step 3.5

Combine $\frac{2}{(v + 1) (v - 1)}$ and $v$ .

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

Step 3.6

Rewrite using the commutative property of multiplication.

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

Step 3.7

Cancel the common factor of $v^{2} - 1$ .

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Step 3.7.1

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

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

Step 3.7.2

Factor $v^{2} - 1$ out of $x (v^{2} - 1)$ .

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

Step 3.7.3

Cancel the common factor.

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

Step 3.7.4

Rewrite the expression.

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

Step 3.8

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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

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

Step 4.2.2

Let $u = (v + 1) (v - 1)$ . 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 4.2.2.1

Let $u = (v + 1) (v - 1)$ . Find $\frac{d u}{d v}$ .

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Step 4.2.2.1.1

Differentiate $(v + 1) (v - 1)$ .

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

Step 4.2.2.1.2

Differentiate using the Product Rule which states that $\frac{d}{d v} [f (v) g (v)]$ is $f (v) \frac{d}{d v} [g (v)] + g (v) \frac{d}{d v} [f (v)]$ where $f (v) = v + 1$ and $g (v) = v - 1$ .

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

Step 4.2.2.1.3

Differentiate.

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Step 4.2.2.1.3.1

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

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

Step 4.2.2.1.3.2

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

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

Step 4.2.2.1.3.3

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

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

Step 4.2.2.1.3.4

Simplify the expression.

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Step 4.2.2.1.3.4.1

Add $1$ and $0$ .

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

Step 4.2.2.1.3.4.2

Multiply $v + 1$ by $1$ .

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

Step 4.2.2.1.3.5

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

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

Step 4.2.2.1.3.6

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

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

Step 4.2.2.1.3.7

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

$v + 1 + (v - 1) (1 + 0)$

Step 4.2.2.1.3.8

Simplify by adding terms.

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Step 4.2.2.1.3.8.1

Add $1$ and $0$ .

$v + 1 + (v - 1) \cdot 1$

Step 4.2.2.1.3.8.2

Multiply $v - 1$ by $1$ .

$v + 1 + v - 1$

Step 4.2.2.1.3.8.3

Add $v$ and $v$ .

$2 v + 1 - 1$

Step 4.2.2.1.3.8.4

Simplify by subtracting numbers.

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Step 4.2.2.1.3.8.4.1

Subtract $1$ from $1$ .

$2 v + 0$

Step 4.2.2.1.3.8.4.2

Add $2 v$ and $0$ .

$2 v$

Step 4.2.2.2

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

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

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

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

Step 4.2.3.2

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

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

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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Step 4.2.5.1

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

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

Step 4.2.5.2

Cancel the common factor of $2$ .

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Step 4.2.5.2.1

Cancel the common factor.

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

Step 4.2.5.2.2

Rewrite the expression.

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

Step 4.2.5.3

Multiply $\int \frac{1}{u} d u$ by $1$ .

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

Step 4.2.6

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

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

Step 4.2.7

Replace all occurrences of $u$ with $(v + 1) (v - 1)$ .

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

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

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

Step 4.3.3

Simplify.

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

Step 4.4

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

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

Step 5

Solve for $v$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Expand $(v + 1) (v - 1)$ using the FOIL Method.

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Step 5.3.1

Apply the distributive property.

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

Step 5.3.2

Apply the distributive property.

$\ln (| v \cdot v + v \cdot - 1 + 1 (v - 1) | | x |) = C$

Step 5.3.3

Apply the distributive property.

$\ln (| v \cdot v + v \cdot - 1 + 1 v + 1 \cdot - 1 | | x |) = C$

Step 5.4

Simplify and combine like terms.

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Step 5.4.1

Simplify each term.

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Step 5.4.1.1

Multiply $v$ by $v$ .

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

Step 5.4.1.2

Move $- 1$ to the left of $v$ .

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

Step 5.4.1.3

Rewrite $- 1 v$ as $- v$ .

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

Step 5.4.1.4

Multiply $v$ by $1$ .

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

Step 5.4.1.5

Multiply $- 1$ by $1$ .

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

Step 5.4.2

Add $- v$ and $v$ .

$\ln (| v^{2} + 0 - 1 | | x |) = C$

Step 5.4.3

Add $v^{2}$ and $0$ .

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

Step 5.5

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

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

Step 5.6

Apply the distributive property.

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

Step 5.7

Rewrite $- 1 x$ as $- x$ .

$\ln (| v^{2} x - x |) = C$

Step 5.8

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

$e^{\ln (| v^{2} x - x |)} = e^{C}$

Step 5.9

Rewrite $\ln (| v^{2} x - x |) = C$ 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 $b^{y} = x$ .

$e^{C} = | v^{2} x - x |$

Step 5.10

Solve for $v$ .

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Step 5.10.1

Rewrite the equation as $| v^{2} x - x | = e^{C}$ .

$| v^{2} x - x | = e^{C}$

Step 5.10.2

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

$v^{2} x - x = \pm e^{C}$

Step 5.10.3

Add $x$ to both sides of the equation.

$v^{2} x = \pm e^{C} + x$

Step 5.10.4

Divide each term in $v^{2} x = \pm e^{C} + x$ by $x$ and simplify.

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Step 5.10.4.1

Divide each term in $v^{2} x = \pm e^{C} + x$ by $x$ .

$\frac{v^{2} x}{x} = \frac{\pm e^{C}}{x} + \frac{x}{x}$

Step 5.10.4.2

Simplify the left side.

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Step 5.10.4.2.1

Cancel the common factor of $x$ .

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Step 5.10.4.2.1.1

Cancel the common factor.

$\frac{v^{2} x}{x} = \frac{\pm e^{C}}{x} + \frac{x}{x}$

Step 5.10.4.2.1.2

Divide $v^{2}$ by $1$ .

$v^{2} = \frac{\pm e^{C}}{x} + \frac{x}{x}$

Step 5.10.4.3

Simplify the right side.

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Step 5.10.4.3.1

Cancel the common factor of $x$ .

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Step 5.10.4.3.1.1

Cancel the common factor.

$v^{2} = \frac{\pm e^{C}}{x} + \frac{x}{x}$

Step 5.10.4.3.1.2

Rewrite the expression.

$v^{2} = \frac{\pm e^{C}}{x} + 1$

Step 5.10.5

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

$v = \pm \sqrt{\frac{\pm e^{C}}{x} + 1}$

Step 5.10.6

Simplify $\pm \sqrt{\frac{\pm e^{C}}{x} + 1}$ .

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Step 5.10.6.1

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

$v = \pm \sqrt{\frac{\pm e^{C}}{x} + \frac{x}{x}}$

Step 5.10.6.2

Combine the numerators over the common denominator.

$v = \pm \sqrt{\frac{\pm e^{C} + x}{x}}$

Step 5.10.6.3

Rewrite $\sqrt{\frac{\pm e^{C} + x}{x}}$ as $\frac{\sqrt{\pm e^{C} + x}}{\sqrt{x}}$ .

$v = \pm \frac{\sqrt{\pm e^{C} + x}}{\sqrt{x}}$

Step 5.10.6.4

Multiply $\frac{\sqrt{\pm e^{C} + x}}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$v = \pm \frac{\sqrt{\pm e^{C} + x}}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}}$

Step 5.10.6.5

Combine and simplify the denominator.

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Step 5.10.6.5.1

Multiply $\frac{\sqrt{\pm e^{C} + x}}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{\sqrt{x} \sqrt{x}}$

Step 5.10.6.5.2

Raise $\sqrt{x}$ to the power of $1$ .

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{{\sqrt{x}}^{1} \sqrt{x}}$

Step 5.10.6.5.3

Raise $\sqrt{x}$ to the power of $1$ .

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{{\sqrt{x}}^{1} {\sqrt{x}}^{1}}$

Step 5.10.6.5.4

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

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{{\sqrt{x}}^{1 + 1}}$

Step 5.10.6.5.5

Add $1$ and $1$ .

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{{\sqrt{x}}^{2}}$

Step 5.10.6.5.6

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

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Step 5.10.6.5.6.1

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

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{{(x^{\frac{1}{2}})}^{2}}$

Step 5.10.6.5.6.2

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

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{x^{\frac{1}{2} \cdot 2}}$

Step 5.10.6.5.6.3

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

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{x^{\frac{2}{2}}}$

Step 5.10.6.5.6.4

Cancel the common factor of $2$ .

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Step 5.10.6.5.6.4.1

Cancel the common factor.

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{x^{\frac{2}{2}}}$

Step 5.10.6.5.6.4.2

Rewrite the expression.

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{x^{1}}$

Step 5.10.6.5.6.5

Simplify.

$v = \pm \frac{\sqrt{\pm e^{C} + x} \sqrt{x}}{x}$

Step 5.10.6.6

Combine using the product rule for radicals.

$v = \pm \frac{\sqrt{(\pm e^{C} + x) x}}{x}$

Step 5.10.6.7

Reorder factors in $\pm \frac{\sqrt{(\pm e^{C} + x) x}}{x}$ .

$v = \pm \frac{\sqrt{x (\pm e^{C} + x)}}{x}$

Step 6

Simplify the constant of integration.

$v = \pm \frac{\sqrt{x (C + x)}}{x}$