Calculus Examples

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

Step 1

Rewrite the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Combine $y$ and $\frac{1}{1 - y}$ .

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

Step 3.3

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

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

Step 3.4

Cancel the common factor of $1 - y$ .

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Step 3.4.1

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

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

Step 3.4.2

Factor $1 - y$ out of $(1 - x^{2}) (1 - y)$ .

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

Step 3.4.3

Cancel the common factor.

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

Step 3.4.4

Rewrite the expression.

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

Step 3.5

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

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

Step 3.6

Simplify the numerator.

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Step 3.6.1

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

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

Step 3.6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Apply the distributive property.

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

Step 4.2.2

Simplify the expression.

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Step 4.2.2.1

Reorder $y$ and $1$ .

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

Step 4.2.2.2

Multiply $y$ by $1$ .

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

Step 4.2.3

Raise $y$ to the power of $1$ .

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

Step 4.2.4

Raise $y$ to the power of $1$ .

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

Step 4.2.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 4.2.6

Add $1$ and $1$ .

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

Step 4.2.7

Reorder $y$ and $y^{2}$ .

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

Step 4.2.8

Reorder $1$ and $- y$ .

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

Step 4.2.9

Divide $y^{2} + y$ by $- y + 1$ .

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Step 4.2.9.1

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

$y$

$1$

$y^{2}$

$y$

$0$

Step 4.2.9.2

Divide the highest order term in the dividend $y^{2}$ by the highest order term in divisor $- y$ .

				-	$y$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$

Step 4.2.9.3

Multiply the new quotient term by the divisor.

				-	$y$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				+	$y^{2}$	-	$y$

Step 4.2.9.4

The expression needs to be subtracted from the dividend, so change all the signs in $y^{2} - y$

				-	$y$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$

Step 4.2.9.5

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

				-	$y$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$
						+	$2 y$

Step 4.2.9.6

Pull the next terms from the original dividend down into the current dividend.

				-	$y$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$
						+	$2 y$	+	$0$

Step 4.2.9.7

Divide the highest order term in the dividend $2 y$ by the highest order term in divisor $- y$ .

				-	$y$	-	$2$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$
						+	$2 y$	+	$0$

Step 4.2.9.8

Multiply the new quotient term by the divisor.

				-	$y$	-	$2$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$
						+	$2 y$	+	$0$
						+	$2 y$	-	$2$

Step 4.2.9.9

The expression needs to be subtracted from the dividend, so change all the signs in $2 y - 2$

				-	$y$	-	$2$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$
						+	$2 y$	+	$0$
						-	$2 y$	+	$2$

Step 4.2.9.10

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

				-	$y$	-	$2$
-	$y$	+	$1$		$y^{2}$	+	$y$	+	$0$
				-	$y^{2}$	+	$y$
						+	$2 y$	+	$0$
						-	$2 y$	+	$2$
								+	$2$

Step 4.2.9.11

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

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

Step 4.2.10

Split the single integral into multiple integrals.

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

Step 4.2.11

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

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

Step 4.2.12

By the Power Rule, the integral of $y$ with respect to $y$ is $\frac{1}{2} y^{2}$ .

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

Step 4.2.13

Apply the constant rule.

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

Step 4.2.14

Combine $\frac{1}{2}$ and $y^{2}$ .

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

Step 4.2.15

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

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

Step 4.2.16

Let $u = - y + 1$ . Then $d u = - d y$ , so $- d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.2.16.1

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

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Step 4.2.16.1.1

Rewrite.

$\frac{- 1}{1}$

Step 4.2.16.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 4.2.16.2

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

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

Step 4.2.17

Move the negative in front of the fraction.

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

Step 4.2.18

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

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

Step 4.2.19

Multiply $- 1$ by $2$ .

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

Step 4.2.20

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

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

Step 4.2.21

Simplify.

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

Step 4.2.22

Replace all occurrences of $u$ with $- y + 1$ .

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Apply the distributive property.

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

Step 4.3.2

Apply the distributive property.

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

Step 4.3.3

Apply the distributive property.

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

Step 4.3.4

Simplify the expression.

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Step 4.3.4.1

Reorder $x$ and $1$ .

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

Step 4.3.4.2

Reorder $x$ and $- 1$ .

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

Step 4.3.4.3

Multiply $1$ by $1$ .

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

Step 4.3.4.4

Multiply $- 1$ by $1$ .

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

Step 4.3.4.5

Multiply $x$ by $1$ .

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

Step 4.3.5

Factor out negative.

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

Step 4.3.6

Raise $x$ to the power of $1$ .

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

Step 4.3.7

Raise $x$ to the power of $1$ .

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

Step 4.3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 4.3.9

Add $1$ and $1$ .

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

Step 4.3.10

Add $- x$ and $x$ .

$- \frac{1}{2} y^{2} - 2 y - 2 \ln (| - y + 1 |) + C_{4} = \int \frac{1 + 0 - x^{2}}{x} d x$

Step 4.3.11

Simplify the expression.

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Step 4.3.11.1

Subtract $x^{2}$ from $0$ .

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

Step 4.3.11.2

Reorder $1$ and $- x^{2}$ .

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

Step 4.3.12

Divide $- x^{2} + 1$ by $x$ .

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Step 4.3.12.1

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

$x$

$0$

$x^{2}$

$0 x$

$1$

Step 4.3.12.2

Divide the highest order term in the dividend $- x^{2}$ by the highest order term in divisor $x$ .

				-	$x$
	$x$	+	$0$	-	$x^{2}$	+	$0 x$	+	$1$

Step 4.3.12.3

Multiply the new quotient term by the divisor.

			-	$x$
$x$	+	$0$	-	$x^{2}$	+	$0 x$	+	$1$
			-	$x^{2}$	+	$0$

Step 4.3.12.4

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{2} + 0$

			-	$x$
$x$	+	$0$	-	$x^{2}$	+	$0 x$	+	$1$
			+	$x^{2}$	-	$0$

Step 4.3.12.5

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

			-	$x$
$x$	+	$0$	-	$x^{2}$	+	$0 x$	+	$1$
			+	$x^{2}$	-	$0$
						$0$

Step 4.3.12.6

Pull the next term from the original dividend down into the current dividend.

			-	$x$
$x$	+	$0$	-	$x^{2}$	+	$0 x$	+	$1$
			+	$x^{2}$	-	$0$
						$0$	+	$1$

Step 4.3.12.7

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

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

Step 4.3.13

Split the single integral into multiple integrals.

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

Step 4.3.14

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

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

Step 4.3.15

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$- \frac{1}{2} y^{2} - 2 y - 2 \ln (| - y + 1 |) + C_{4} = - (\frac{1}{2} x^{2} + C_{5}) + \int \frac{1}{x} d x$

Step 4.3.16

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

$- \frac{1}{2} y^{2} - 2 y - 2 \ln (| - y + 1 |) + C_{4} = - (\frac{1}{2} x^{2} + C_{5}) + \ln (| x |) + C_{6}$

Step 4.3.17

Simplify.

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Step 4.3.17.1

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

$- \frac{1}{2} y^{2} - 2 y - 2 \ln (| - y + 1 |) + C_{4} = - (\frac{x^{2}}{2} + C_{5}) + \ln (| x |) + C_{6}$

Step 4.3.17.2

Simplify.

$- \frac{1}{2} y^{2} - 2 y - 2 \ln (| - y + 1 |) + C_{4} = - \frac{1}{2} x^{2} + \ln (| x |) + C_{7}$

Step 4.4

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

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