Calculus Examples

$x \cdot y + x + \frac{d y}{d x} \cdot y + \frac{d y}{d x} \cdot y \cdot x = 0$

Step 1

Separate the variables.

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Step 1.1

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

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Step 1.1.1

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

$x y + x + \frac{d y}{d x} y + \frac{d y}{d x} y x = 0$

Step 1.1.2

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

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Step 1.1.2.1

Subtract $x y$ from both sides of the equation.

$x + \frac{d y}{d x} y + \frac{d y}{d x} y x = - x y$

Step 1.1.2.2

Subtract $x$ from both sides of the equation.

$\frac{d y}{d x} y + \frac{d y}{d x} y x = - x y - x$

Step 1.1.3

Factor $\frac{d y}{d x} y$ out of $\frac{d y}{d x} y + \frac{d y}{d x} y x$ .

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Step 1.1.3.1

Factor $\frac{d y}{d x} y$ out of $\frac{d y}{d x} y$ .

$\frac{d y}{d x} y \cdot 1 + \frac{d y}{d x} y x = - x y - x$

Step 1.1.3.2

Factor $\frac{d y}{d x} y$ out of $\frac{d y}{d x} y x$ .

$\frac{d y}{d x} y \cdot 1 + \frac{d y}{d x} y x = - x y - x$

Step 1.1.3.3

Factor $\frac{d y}{d x} y$ out of $\frac{d y}{d x} y \cdot 1 + \frac{d y}{d x} y x$ .

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

Step 1.1.4

Divide each term in $\frac{d y}{d x} y (1 + x) = - x y - x$ by $y (1 + x)$ and simplify.

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Step 1.1.4.1

Divide each term in $\frac{d y}{d x} y (1 + x) = - x y - x$ by $y (1 + x)$ .

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

Step 1.1.4.2

Simplify the left side.

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Step 1.1.4.2.1

Cancel the common factor of $y$ .

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Step 1.1.4.2.1.1

Cancel the common factor.

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

Step 1.1.4.2.1.2

Rewrite the expression.

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

Step 1.1.4.2.2

Cancel the common factor of $1 + x$ .

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Step 1.1.4.2.2.1

Cancel the common factor.

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

Step 1.1.4.2.2.2

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

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

Step 1.1.4.3

Simplify the right side.

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Step 1.1.4.3.1

Simplify each term.

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Step 1.1.4.3.1.1

Cancel the common factor of $y$ .

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Step 1.1.4.3.1.1.1

Cancel the common factor.

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

Step 1.1.4.3.1.1.2

Rewrite the expression.

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

Step 1.1.4.3.1.2

Move the negative in front of the fraction.

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

Step 1.1.4.3.1.3

Move the negative in front of the fraction.

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

Step 1.1.4.3.2

To write $- \frac{x}{1 + x}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

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

Step 1.1.4.3.3

Write each expression with a common denominator of $(1 + x) y$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.1.4.3.3.1

Multiply $\frac{x}{1 + x}$ by $\frac{y}{y}$ .

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

Step 1.1.4.3.3.2

Reorder the factors of $(1 + x) y$ .

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

Step 1.1.4.3.4

Combine the numerators over the common denominator.

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

Step 1.1.4.3.5

Factor $x$ out of $- x y - x$ .

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Step 1.1.4.3.5.1

Factor $x$ out of $- x y$ .

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

Step 1.1.4.3.5.2

Factor $x$ out of $- x$ .

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

Step 1.1.4.3.5.3

Factor $x$ out of $x (- y) + x \cdot - 1$ .

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

Step 1.1.4.3.6

Factor $- 1$ out of $- y$ .

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

Step 1.1.4.3.7

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 1.1.4.3.8

Factor $- 1$ out of $- (y) - 1 (1)$ .

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

Step 1.1.4.3.9

Simplify the expression.

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Step 1.1.4.3.9.1

Rewrite $- (y + 1)$ as $- 1 (y + 1)$ .

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

Step 1.1.4.3.9.2

Move the negative in front of the fraction.

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

Step 1.2

Regroup factors.

$\frac{d y}{d x} = - \frac{x}{1 + x} \cdot \frac{y + 1}{y}$

Step 1.3

Multiply both sides by $\frac{y}{y + 1}$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.4.2

Multiply $\frac{x}{1 + x}$ by $\frac{y + 1}{y}$ .

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

Step 1.4.3

Cancel the common factor of $y$ .

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Step 1.4.3.1

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

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

Step 1.4.3.2

Factor $y$ out of $- y$ .

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

Step 1.4.3.3

Factor $y$ out of $(1 + x) y$ .

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

Step 1.4.3.4

Cancel the common factor.

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

Step 1.4.3.5

Rewrite the expression.

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

Step 1.4.4

Cancel the common factor of $y + 1$ .

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Step 1.4.4.1

Factor $y + 1$ out of $x (y + 1)$ .

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

Step 1.4.4.2

Cancel the common factor.

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

Step 1.4.4.3

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Divide $y$ by $y + 1$ .

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Step 2.2.1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y$

$1$

$y$

$0$

Step 2.2.1.2

Divide the highest order term in the dividend $y$ by the highest order term in divisor $y$ .

					$1$
	$y$	+	$1$		$y$	+	$0$

Step 2.2.1.3

Multiply the new quotient term by the divisor.

				$1$
$y$	+	$1$		$y$	+	$0$
			+	$y$	+	$1$

Step 2.2.1.4

The expression needs to be subtracted from the dividend, so change all the signs in $y + 1$

				$1$
$y$	+	$1$		$y$	+	$0$
			-	$y$	-	$1$

Step 2.2.1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$y$	+	$1$		$y$	+	$0$
			-	$y$	-	$1$
					-	$1$

Step 2.2.1.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

Apply the constant rule.

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

Step 2.2.4

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

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

Step 2.2.5

Let $u_{1} = y + 1$ . Then $d u_{1} = d y$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 2.2.5.1

Let $u_{1} = y + 1$ . Find $\frac{d u_{1}}{d y}$ .

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Step 2.2.5.1.1

Differentiate $y + 1$ .

$\frac{d}{d y} [y + 1]$

Step 2.2.5.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [1]$

Step 2.2.5.1.3

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

$1 + \frac{d}{d y} [1]$

Step 2.2.5.1.4

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

$1 + 0$

Step 2.2.5.1.5

Add $1$ and $0$ .

$1$

Step 2.2.5.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

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

Step 2.2.6

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

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

Step 2.2.7

Simplify.

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

Step 2.2.8

Replace all occurrences of $u_{1}$ with $y + 1$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Reorder $1$ and $x$ .

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

Step 2.3.3

Divide $x$ by $x + 1$ .

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Step 2.3.3.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x$

$0$

Step 2.3.3.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

					$1$
	$x$	+	$1$		$x$	+	$0$

Step 2.3.3.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$1$		$x$	+	$0$
			+	$x$	+	$1$

Step 2.3.3.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 1$

				$1$
$x$	+	$1$		$x$	+	$0$
			-	$x$	-	$1$

Step 2.3.3.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$x$	+	$1$		$x$	+	$0$
			-	$x$	-	$1$
					-	$1$

Step 2.3.3.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 2.3.4

Split the single integral into multiple integrals.

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

Step 2.3.5

Apply the constant rule.

$y - \ln (| y + 1 |) + C_{3} = - (x + C_{4} + \int - \frac{1}{x + 1} d x)$

Step 2.3.6

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

$y - \ln (| y + 1 |) + C_{3} = - (x + C_{4} - \int \frac{1}{x + 1} d x)$

Step 2.3.7

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.7.1

Let $u_{2} = x + 1$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.7.1.1

Differentiate $x + 1$ .

$\frac{d}{d x} [x + 1]$

Step 2.3.7.1.2

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

$\frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 2.3.7.1.3

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

$1 + \frac{d}{d x} [1]$

Step 2.3.7.1.4

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

$1 + 0$

Step 2.3.7.1.5

Add $1$ and $0$ .

$1$

Step 2.3.7.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$y - \ln (| y + 1 |) + C_{3} = - (x + C_{4} - \int \frac{1}{u_{2}} d u_{2})$

Step 2.3.8

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$y - \ln (| y + 1 |) + C_{3} = - (x + C_{4} - (\ln (| u_{2} |) + C_{5}))$

Step 2.3.9

Simplify.

$y - \ln (| y + 1 |) + C_{3} = - (x - \ln (| u_{2} |)) + C_{6}$

Step 2.3.10

Replace all occurrences of $u_{2}$ with $x + 1$ .

$y - \ln (| y + 1 |) + C_{3} = - (x - \ln (| x + 1 |)) + C_{6}$

Step 2.3.11

Simplify.

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Step 2.3.11.1

Apply the distributive property.

$y - \ln (| y + 1 |) + C_{3} = - x - - \ln (| x + 1 |) + C_{6}$

Step 2.3.11.2

Multiply $- - \ln (| x + 1 |)$ .

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Step 2.3.11.2.1

Multiply $- 1$ by $- 1$ .

$y - \ln (| y + 1 |) + C_{3} = - x + 1 \ln (| x + 1 |) + C_{6}$

Step 2.3.11.2.2

Multiply $\ln (| x + 1 |)$ by $1$ .

$y - \ln (| y + 1 |) + C_{3} = - x + \ln (| x + 1 |) + C_{6}$

Step 2.4

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

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