Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $1 + x$ .

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Multiply $\frac{1}{y}$ by $1 - y$ .

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

Step 3.3

Rewrite using the commutative property of multiplication.

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

Step 3.4

Cancel the common factor of $y$ .

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

Factor $y$ out of $x y$ .

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

Step 3.4.4

Cancel the common factor.

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

Step 3.4.5

Rewrite the expression.

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

Step 3.5

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

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Split the fraction into multiple fractions.

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

Step 4.2.2

Split the single integral into multiple integrals.

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

Step 4.2.3

Cancel the common factor of $y$ .

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Step 4.2.3.1

Cancel the common factor.

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

Step 4.2.3.2

Divide $- 1$ by $1$ .

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

Step 4.2.4

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

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

Step 4.2.5

Apply the constant rule.

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

Step 4.2.6

Simplify.

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

Step 4.2.7

Reorder terms.

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

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.

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

Step 4.3.2

Reorder $1$ and $x$ .

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

Step 4.3.3

Divide $x$ by $x + 1$ .

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Step 4.3.3.1

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 4.3.3.2

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

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

Step 4.3.3.3

Multiply the new quotient term by the divisor.

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

Step 4.3.3.4

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 4.3.3.5

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 4.3.3.6

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

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

Step 4.3.4

Split the single integral into multiple integrals.

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

Step 4.3.5

Apply the constant rule.

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

Step 4.3.6

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

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

Step 4.3.7

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

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Step 4.3.7.1

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

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Step 4.3.7.1.1

Differentiate $x + 1$ .

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

Step 4.3.7.1.2

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 4.3.7.1.3

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 4.3.7.1.4

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

$1 + 0$

Step 4.3.7.1.5

Add $1$ and $0$ .

$1$

Step 4.3.7.2

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

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

Step 4.3.8

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

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

Step 4.3.9

Simplify.

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

Step 4.3.10

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

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

Step 4.3.11

Simplify.

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Step 4.3.11.1

Apply the distributive property.

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

Step 4.3.11.2

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

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Step 4.3.11.2.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.11.2.2

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

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

Step 4.4

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

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