Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{1 - 2 v}$ .

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

Step 1.2

Cancel the common factor of $1 - 2 v$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = 1 - 2 v$ . Then $d u = - 2 d v$ , so $- \frac{1}{2} d u = d v$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = 1 - 2 v$ . Find $\frac{d u}{d v}$ .

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Step 2.2.1.1.1

Differentiate $1 - 2 v$ .

$\frac{d}{d v} [1 - 2 v]$

Step 2.2.1.1.2

Differentiate.

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Step 2.2.1.1.2.1

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

$\frac{d}{d v} [1] + \frac{d}{d v} [- 2 v]$

Step 2.2.1.1.2.2

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

$0 + \frac{d}{d v} [- 2 v]$

Step 2.2.1.1.3

Evaluate $\frac{d}{d v} [- 2 v]$ .

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Step 2.2.1.1.3.1

Since $- 2$ is constant with respect to $v$ , the derivative of $- 2 v$ with respect to $v$ is $- 2 \frac{d}{d v} [v]$ .

$0 - 2 \frac{d}{d v} [v]$

Step 2.2.1.1.3.2

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

$0 - 2 \cdot 1$

Step 2.2.1.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 2.2.1.1.4

Subtract $2$ from $0$ .

$- 2$

Step 2.2.1.2

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

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

Step 2.2.2

Simplify.

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Step 2.2.2.1

Move the negative in front of the fraction.

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

Simplify.

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

Step 2.2.7

Replace all occurrences of $u$ with $1 - 2 v$ .

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

Step 2.3

Apply the constant rule.

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

Step 2.4

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

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

Step 3

Solve for $v$ .

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Step 3.1

Multiply both sides of the equation by $- 2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $- 2 (- \frac{1}{2} \ln (| 1 - 2 v |))$ .

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

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

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

Step 3.2.1.1.2.2

Factor $2$ out of $- 2$ .

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

Step 3.2.1.1.2.3

Cancel the common factor.

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

Step 3.2.1.1.2.4

Rewrite the expression.

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

Step 3.2.1.1.3

Multiply.

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Step 3.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.2.1.1.3.2

Multiply $\ln (| 1 - 2 v |)$ by $1$ .

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

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

Step 3.3

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

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

Step 3.4

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

$e^{- 2 x - 2 C} = | 1 - 2 v |$

Step 3.5

Solve for $v$ .

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Step 3.5.1

Rewrite the equation as $| 1 - 2 v | = e^{- 2 x - 2 C}$ .

$| 1 - 2 v | = e^{- 2 x - 2 C}$

Step 3.5.2

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

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

Step 3.5.3

Subtract $1$ from both sides of the equation.

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

Step 3.5.4

Divide each term in $- 2 v = \pm e^{- 2 x - 2 C} - 1$ by $- 2$ and simplify.

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Step 3.5.4.1

Divide each term in $- 2 v = \pm e^{- 2 x - 2 C} - 1$ by $- 2$ .

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

Step 3.5.4.2

Simplify the left side.

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Step 3.5.4.2.1

Cancel the common factor of $- 2$ .

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Step 3.5.4.2.1.1

Cancel the common factor.

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

Step 3.5.4.2.1.2

Divide $v$ by $1$ .

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

Step 3.5.4.3

Simplify the right side.

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Step 3.5.4.3.1

Simplify each term.

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Step 3.5.4.3.1.1

Simplify $\frac{\pm e^{- 2 x - 2 C}}{- 2}$ .

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

Step 3.5.4.3.1.2

Dividing two negative values results in a positive value.

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

Step 3.5.4.3.2

Combine the numerators over the common denominator.

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Rewrite $e^{- 2 x + C}$ as $e^{- 2 x} e^{C}$ .

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

Step 4.3

Reorder $e^{- 2 x}$ and $e^{C}$ .

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

Step 4.4

Combine constants with the plus or minus.

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