Calculus Examples

$(x + 2) d x - x d y = 0$

Step 1

Subtract $(x + 2) d x$ from both sides of the equation.

$- x d y = - (x + 2) d x$

Step 2

Multiply both sides by $\frac{1}{x}$ .

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

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

Step 3.2

Cancel the common factor of $x$ .

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Step 3.2.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 3.3

Rewrite using the commutative property of multiplication.

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

Step 3.4

Apply the distributive property.

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

Step 3.5

Cancel the common factor of $x$ .

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Step 3.5.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Step 3.5.2

Cancel the common factor.

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

Step 3.5.3

Rewrite the expression.

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

Step 3.6

Multiply $- \frac{1}{x} \cdot 2$ .

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Step 3.6.1

Multiply $2$ by $- 1$ .

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

Step 3.6.2

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

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

Step 3.7

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Apply the constant rule.

$- y + C_{1} = \int - 1 - \frac{2}{x} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

Split the single integral into multiple integrals.

$- y + C_{1} = \int - 1 d x + \int - \frac{2}{x} d x$

Step 4.3.2

Apply the constant rule.

$- y + C_{1} = - x + C_{2} + \int - \frac{2}{x} d x$

Step 4.3.3

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

$- y + C_{1} = - x + C_{2} - \int \frac{2}{x} d x$

Step 4.3.4

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

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

Step 4.3.5

Multiply $2$ by $- 1$ .

$- y + C_{1} = - x + C_{2} - 2 \int \frac{1}{x} d x$

Step 4.3.6

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

$- y + C_{1} = - x + C_{2} - 2 (\ln (| x |) + C_{3})$

Step 4.3.7

Simplify.

$- y + C_{1} = - x - 2 \ln (| x |) + C_{4}$

Step 4.4

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

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

Step 5

Divide each term in $- y = - x - 2 \ln (| x |) + C$ by $- 1$ and simplify.

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Step 5.1

Divide each term in $- y = - x - 2 \ln (| x |) + C$ by $- 1$ .

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Dividing two negative values results in a positive value.

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

Step 5.2.2

Divide $y$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Simplify each term.

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Step 5.3.1.1

Dividing two negative values results in a positive value.

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

Step 5.3.1.2

Divide $x$ by $1$ .

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

Step 5.3.1.3

Move the negative one from the denominator of $\frac{- 2 \ln (| x |)}{- 1}$ .

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

Step 5.3.1.4

Rewrite $- 1 \cdot (- 2 \ln (| x |))$ as $- (- 2 \ln (| x |))$ .

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

Step 5.3.1.5

Multiply $- (- 2 \ln (| x |))$ .

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Step 5.3.1.5.1

Simplify $- 2 \ln (| x |)$ by moving $2$ inside the logarithm.

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

Step 5.3.1.5.2

Multiply $- 1$ by $- 1$ .

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

Step 5.3.1.5.3

Multiply $\ln ({| x |}^{2})$ by $1$ .

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

Step 5.3.1.6

Remove the absolute value in ${| x |}^{2}$ because exponentiations with even powers are always positive.

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

Step 5.3.1.7

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 5.3.1.8

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 6

Simplify the constant of integration.

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