Calculus Examples

$x (v^{2} - 1) d v + (v^{3} - 3 v) d x = 0$

Step 1

Subtract $(v^{3} - 3 v) d x$ from both sides of the equation.

$x (v^{2} - 1) d v = - (v^{3} - 3 v) d x$

Step 2

Multiply both sides by $\frac{1}{x (v^{3} - 3 v)}$ .

$\frac{1}{x (v^{3} - 3 v)} (x (v^{2} - 1)) d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Cancel the common factor.

$\frac{1}{x (v^{3} - 3 v)} (x (v^{2} - 1)) d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3.1.2

Rewrite the expression.

$\frac{1}{v^{3} - 3 v} (v^{2} - 1) d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3.2

Factor $v$ out of $v^{3} - 3 v$ .

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Step 3.2.1

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

$\frac{1}{v \cdot v^{2} - 3 v} (v^{2} - 1) d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3.2.2

Factor $v$ out of $- 3 v$ .

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

Step 3.2.3

Factor $v$ out of $v \cdot v^{2} + v \cdot - 3$ .

$\frac{1}{v (v^{2} - 3)} (v^{2} - 1) d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3.3

Multiply $\frac{1}{v (v^{2} - 3)}$ by $v^{2} - 1$ .

$\frac{v^{2} - 1}{v (v^{2} - 3)} d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3.4

Simplify the numerator.

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Step 3.4.1

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

$\frac{v^{2} - 1^{2}}{v (v^{2} - 3)} d v = \frac{1}{x (v^{3} - 3 v)} (- (v^{3} - 3 v)) d x$

Step 3.4.2

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 $b = 1$ .

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

Step 3.5

Rewrite using the commutative property of multiplication.

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

Step 3.6

Cancel the common factor of $v^{3} - 3 v$ .

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Step 3.6.1

Move the leading negative in $- \frac{1}{x (v^{3} - 3 v)}$ into the numerator.

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

Step 3.6.2

Factor $v^{3} - 3 v$ out of $x (v^{3} - 3 v)$ .

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

Step 3.6.3

Cancel the common factor.

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

Step 3.6.4

Rewrite the expression.

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

Step 3.7

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int \frac{(v + 1) (v - 1)}{v (v^{2} - 3)} d v = \int - \frac{1}{x} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Write the fraction using partial fraction decomposition.

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Step 4.2.1.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.2.1.1.1

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.

$\frac{A}{v} + \frac{B v + C}{v^{2} - 3}$

Step 4.2.1.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $v (v^{2} - 3)$ .

$\frac{(v + 1) (v - 1) (v (v^{2} - 3))}{v (v^{2} - 3)} = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.3

Reduce the expression by cancelling the common factors.

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Step 4.2.1.1.3.1

Cancel the common factor of $v$ .

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Step 4.2.1.1.3.1.1

Cancel the common factor.

$\frac{(v + 1) (v - 1) (v (v^{2} - 3))}{v (v^{2} - 3)} = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.3.1.2

Rewrite the expression.

$\frac{(v + 1) (v - 1) (v^{2} - 3)}{v^{2} - 3} = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.3.2

Cancel the common factor of $v^{2} - 3$ .

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Step 4.2.1.1.3.2.1

Cancel the common factor.

$\frac{(v + 1) (v - 1) (v^{2} - 3)}{v^{2} - 3} = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.3.2.2

Divide $(v + 1) (v - 1)$ by $1$ .

$(v + 1) (v - 1) = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.4

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

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Step 4.2.1.1.4.1

Apply the distributive property.

$v (v - 1) + 1 (v - 1) = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.4.2

Apply the distributive property.

$v \cdot v + v \cdot - 1 + 1 (v - 1) = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.4.3

Apply the distributive property.

$v \cdot v + v \cdot - 1 + 1 v + 1 \cdot - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5

Simplify and combine like terms.

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Step 4.2.1.1.5.1

Simplify each term.

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Step 4.2.1.1.5.1.1

Multiply $v$ by $v$ .

$v^{2} + v \cdot - 1 + 1 v + 1 \cdot - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5.1.2

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

$v^{2} - 1 \cdot v + 1 v + 1 \cdot - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5.1.3

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

$v^{2} - v + 1 v + 1 \cdot - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5.1.4

Multiply $v$ by $1$ .

$v^{2} - v + v + 1 \cdot - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5.1.5

Multiply $- 1$ by $1$ .

$v^{2} - v + v - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5.2

Add $- v$ and $v$ .

$v^{2} + 0 - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.5.3

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

$v^{2} - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.6

Simplify each term.

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Step 4.2.1.1.6.1

Cancel the common factor of $v$ .

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Step 4.2.1.1.6.1.1

Cancel the common factor.

$v^{2} - 1 = \frac{A (v (v^{2} - 3))}{v} + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.6.1.2

Divide $A (v^{2} - 3)$ by $1$ .

$v^{2} - 1 = A (v^{2} - 3) + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.6.2

Apply the distributive property.

$v^{2} - 1 = A v^{2} + A \cdot - 3 + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.6.3

Move $- 3$ to the left of $A$ .

$v^{2} - 1 = A v^{2} - 3 \cdot A + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.6.4

Cancel the common factor of $v^{2} - 3$ .

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Step 4.2.1.1.6.4.1

Cancel the common factor.

$v^{2} - 1 = A v^{2} - 3 A + \frac{(B v + C) (v (v^{2} - 3))}{v^{2} - 3}$

Step 4.2.1.1.6.4.2

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

$v^{2} - 1 = A v^{2} - 3 A + (B v + C) (v)$

Step 4.2.1.1.6.5

Apply the distributive property.

$v^{2} - 1 = A v^{2} - 3 A + B v \cdot v + C v$

Step 4.2.1.1.6.6

Multiply $v$ by $v$ by adding the exponents.

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Step 4.2.1.1.6.6.1

Move $v$ .

$v^{2} - 1 = A v^{2} - 3 A + B (v \cdot v) + C v$

Step 4.2.1.1.6.6.2

Multiply $v$ by $v$ .

$v^{2} - 1 = A v^{2} - 3 A + B v^{2} + C v$

Step 4.2.1.1.7

Move $- 3 A$ .

$v^{2} - 1 = A v^{2} + B v^{2} + C v - 3 A$

Step 4.2.1.2

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

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Step 4.2.1.2.1

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.

$1 = A + B$

Step 4.2.1.2.2

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 4.2.1.2.3

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.

$- 1 = - 3 A$

Step 4.2.1.2.4

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

$1 = A + B$

$0 = C$

$- 1 = - 3 A$

$1 = A + B$

$0 = C$

$- 1 = - 3 A$

Step 4.2.1.3

Solve the system of equations.

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Step 4.2.1.3.1

Rewrite the equation as $C = 0$ .

$C = 0$

$1 = A + B$

$- 1 = - 3 A$

Step 4.2.1.3.2

Replace all occurrences of $C$ with $0$ in each equation.

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Step 4.2.1.3.2.1

Rewrite the equation as $- 3 A = - 1$ .

$- 3 A = - 1$

$C = 0$

$1 = A + B$

Step 4.2.1.3.2.2

Divide each term in $- 3 A = - 1$ by $- 3$ and simplify.

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Step 4.2.1.3.2.2.1

Divide each term in $- 3 A = - 1$ by $- 3$ .

$\frac{- 3 A}{- 3} = \frac{- 1}{- 3}$

$C = 0$

$1 = A + B$

Step 4.2.1.3.2.2.2

Simplify the left side.

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Step 4.2.1.3.2.2.2.1

Cancel the common factor of $- 3$ .

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Step 4.2.1.3.2.2.2.1.1

Cancel the common factor.

$\frac{- 3 A}{- 3} = \frac{- 1}{- 3}$

$C = 0$

$1 = A + B$

Step 4.2.1.3.2.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 1}{- 3}$

$C = 0$

$1 = A + B$

$A = \frac{- 1}{- 3}$

$C = 0$

$1 = A + B$

$A = \frac{- 1}{- 3}$

$C = 0$

$1 = A + B$

Step 4.2.1.3.2.2.3

Simplify the right side.

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Step 4.2.1.3.2.2.3.1

Dividing two negative values results in a positive value.

$A = \frac{1}{3}$

$C = 0$

$1 = A + B$

$A = \frac{1}{3}$

$C = 0$

$1 = A + B$

$A = \frac{1}{3}$

$C = 0$

$1 = A + B$

$A = \frac{1}{3}$

$C = 0$

$1 = A + B$

Step 4.2.1.3.3

Replace all occurrences of $A$ with $\frac{1}{3}$ in each equation.

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Step 4.2.1.3.3.1

Replace all occurrences of $A$ in $1 = A + B$ with $\frac{1}{3}$ .

$1 = (\frac{1}{3}) + B$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.3.2

Simplify the right side.

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Step 4.2.1.3.3.2.1

Remove parentheses.

$1 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$C = 0$

$1 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$C = 0$

$1 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.4

Solve for $B$ in $1 = \frac{1}{3} + B$ .

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Step 4.2.1.3.4.1

Rewrite the equation as $\frac{1}{3} + B = 1$ .

$\frac{1}{3} + B = 1$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.4.2

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

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Step 4.2.1.3.4.2.1

Subtract $\frac{1}{3}$ from both sides of the equation.

$B = 1 - \frac{1}{3}$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.4.2.2

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

$B = \frac{3}{3} - \frac{1}{3}$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.4.2.3

Combine the numerators over the common denominator.

$B = \frac{3 - 1}{3}$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.4.2.4

Subtract $1$ from $3$ .

$B = \frac{2}{3}$

$A = \frac{1}{3}$

$C = 0$

$B = \frac{2}{3}$

$A = \frac{1}{3}$

$C = 0$

$B = \frac{2}{3}$

$A = \frac{1}{3}$

$C = 0$

Step 4.2.1.3.5

Solve the system of equations.

$B = \frac{2}{3} A = \frac{1}{3} C = 0$

Step 4.2.1.3.6

List all of the solutions.

$B = \frac{2}{3}, A = \frac{1}{3}, C = 0$

Step 4.2.1.4

Replace each of the partial fraction coefficients in $\frac{A}{v} + \frac{B v + C}{v^{2} - 3}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{3}}{v} + \frac{\frac{2}{3 v} + 0}{v^{2} - 3}$

Step 4.2.1.5

Simplify.

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Step 4.2.1.5.1

Remove parentheses.

$\frac{\frac{1}{3}}{v} + \frac{(\frac{2}{3}) v + 0}{v^{2} - 3}$

Step 4.2.1.5.2

Simplify the numerator.

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Step 4.2.1.5.2.1

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

$\frac{\frac{1}{3}}{v} + \frac{\frac{2 v}{3} + 0}{v^{2} - 3}$

Step 4.2.1.5.2.2

Add $\frac{2 v}{3}$ and $0$ .

$\frac{\frac{1}{3}}{v} + \frac{\frac{2 v}{3}}{v^{2} - 3}$

Step 4.2.1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{\frac{1}{3}}{v} + \frac{2 v}{3} \cdot \frac{1}{v^{2} - 3}$

Step 4.2.1.5.4

Multiply $\frac{2 v}{3}$ by $\frac{1}{v^{2} - 3}$ .

$\frac{\frac{1}{3}}{v} + \frac{2 v}{3 (v^{2} - 3)}$

Step 4.2.1.5.5

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3} \cdot \frac{1}{v} + \frac{2 v}{3 (v^{2} - 3)}$

Step 4.2.1.5.6

Multiply $\frac{1}{3}$ by $\frac{1}{v}$ .

$\int \frac{1}{3 v} + \frac{2 v}{3 (v^{2} - 3)} d v = \int - \frac{1}{x} d x$

Step 4.2.2

Split the single integral into multiple integrals.

$\int \frac{1}{3 v} d v + \int \frac{2 v}{3 (v^{2} - 3)} d v = \int - \frac{1}{x} d x$

Step 4.2.3

Since $\frac{1}{3}$ is constant with respect to $v$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int \frac{1}{v} d v + \int \frac{2 v}{3 (v^{2} - 3)} d v = \int - \frac{1}{x} d x$

Step 4.2.4

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

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

Step 4.2.5

Since $\frac{2}{3}$ is constant with respect to $v$ , move $\frac{2}{3}$ out of the integral.

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

Step 4.2.6

Let $u = v^{2} - 3$ . 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 4.2.6.1

Let $u = v^{2} - 3$ . Find $\frac{d u}{d v}$ .

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Step 4.2.6.1.1

Differentiate $v^{2} - 3$ .

$\frac{d}{d v} [v^{2} - 3]$

Step 4.2.6.1.2

By the Sum Rule, the derivative of $v^{2} - 3$ with respect to $v$ is $\frac{d}{d v} [v^{2}] + \frac{d}{d v} [- 3]$ .

$\frac{d}{d v} [v^{2}] + \frac{d}{d v} [- 3]$

Step 4.2.6.1.3

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

$2 v + \frac{d}{d v} [- 3]$

Step 4.2.6.1.4

Since $- 3$ is constant with respect to $v$ , the derivative of $- 3$ with respect to $v$ is $0$ .

$2 v + 0$

Step 4.2.6.1.5

Add $2 v$ and $0$ .

$2 v$

Step 4.2.6.2

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

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

Step 4.2.7

Simplify.

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Step 4.2.7.1

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

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

Step 4.2.7.2

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

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

Step 4.2.8

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

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

Step 4.2.9

Simplify.

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Step 4.2.9.1

Multiply $\frac{1}{2}$ by $\frac{2}{3}$ .

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

Step 4.2.9.2

Multiply $2$ by $3$ .

$\frac{1}{3} (\ln (| v |) + C_{1}) + \frac{2}{6} \int \frac{1}{u} d u = \int - \frac{1}{x} d x$

Step 4.2.9.3

Cancel the common factor of $2$ and $6$ .

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Step 4.2.9.3.1

Factor $2$ out of $2$ .

$\frac{1}{3} (\ln (| v |) + C_{1}) + \frac{2 (1)}{6} \int \frac{1}{u} d u = \int - \frac{1}{x} d x$

Step 4.2.9.3.2

Cancel the common factors.

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Step 4.2.9.3.2.1

Factor $2$ out of $6$ .

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

Step 4.2.9.3.2.2

Cancel the common factor.

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

Step 4.2.9.3.2.3

Rewrite the expression.

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

Step 4.2.10

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

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

Step 4.2.11

Simplify.

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

Step 4.2.12

Replace all occurrences of $u$ with $v^{2} - 3$ .

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

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.

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

Step 4.3.2

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

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

Step 4.3.3

Simplify.

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

Step 4.4

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

$\frac{1}{3} \ln (| v |) + \frac{1}{3} \ln (| v^{2} - 3 |) = - \ln (| x |) + C$