Calculus Examples

$x (y d) x + (x + 1) d y = 0$

Step 1

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

$(x + 1) d y = - x y d x$

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x + 1$ .

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Factor $y$ out of $x y$ .

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

Step 3.3.4

Cancel the common factor.

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

Step 3.3.5

Rewrite the expression.

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

Step 3.4

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

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

Step 3.5

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

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.

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

Step 4.3.2

Divide $x$ by $x + 1$ .

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Step 4.3.2.1

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 4.3.2.2

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

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

Step 4.3.2.3

Multiply the new quotient term by the divisor.

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

Step 4.3.2.4

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 4.3.2.5

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 4.3.2.6

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

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

Step 4.3.3

Split the single integral into multiple integrals.

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

Step 4.3.4

Apply the constant rule.

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

Step 4.3.5

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

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

Step 4.3.6

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.6.1

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

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Step 4.3.6.1.1

Differentiate $x + 1$ .

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

Step 4.3.6.1.2

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 4.3.6.1.3

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 4.3.6.1.4

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

$1 + 0$

Step 4.3.6.1.5

Add $1$ and $0$ .

$1$

Step 4.3.6.2

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

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

Step 4.3.7

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

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

Step 4.3.8

Simplify.

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

Step 4.3.9

Replace all occurrences of $u$ with $x + 1$ .

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

Step 4.3.10

Simplify.

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Step 4.3.10.1

Apply the distributive property.

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

Step 4.3.10.2

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

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Step 4.3.10.2.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.10.2.2

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

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

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

Step 5.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 5.3

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

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

Step 5.4

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

$e^{- x + C} = \frac{| y |}{| x + 1 |}$

Step 5.5

Solve for $y$ .

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Step 5.5.1

Rewrite the equation as $\frac{| y |}{| x + 1 |} = e^{- x + C}$ .

$\frac{| y |}{| x + 1 |} = e^{- x + C}$

Step 5.5.2

Multiply both sides by $| x + 1 |$ .

$\frac{| y |}{| x + 1 |} | x + 1 | = e^{- x + C} | x + 1 |$

Step 5.5.3

Simplify the left side.

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Step 5.5.3.1

Cancel the common factor of $| x + 1 |$ .

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Step 5.5.3.1.1

Cancel the common factor.

$\frac{| y |}{| x + 1 |} | x + 1 | = e^{- x + C} | x + 1 |$

Step 5.5.3.1.2

Rewrite the expression.

$| y | = e^{- x + C} | x + 1 |$

Step 5.5.4

Solve for $y$ .

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Step 5.5.4.1

Reorder factors in $e^{- x + C} | x + 1 |$ .

$| y | = | x + 1 | e^{- x + C}$

Step 5.5.4.2

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

$y = \pm | x + 1 | e^{- x + C}$

Step 6

Group the constant terms together.

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Step 6.1

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

$y = \pm | x + 1 | (e^{- x} e^{C})$

Step 6.2

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

$y = \pm | x + 1 | (e^{C} e^{- x})$

Step 6.3

Combine constants with the plus or minus.

$y = C | x + 1 | e^{- x}$