Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

Multiply both sides by $\frac{1}{y^{2}}$ .

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

Step 1.3

Cancel the common factor of $y^{2}$ .

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Step 1.3.1

Factor $y^{2}$ out of $\frac{x^{2}}{1 - x} y^{2}$ .

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

Step 1.3.2

Cancel the common factor.

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

Step 1.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Apply basic rules of exponents.

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Step 2.2.1.1

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.2.1.2

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 2.2.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.2.2

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

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

Step 2.2.3

Rewrite $- y^{- 1} + C_{1}$ as $- \frac{1}{y} + C_{1}$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Reorder $1$ and $- x$ .

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

Step 2.3.2

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

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Step 2.3.2.1

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

$x$

$1$

$x^{2}$

$0 x$

$0$

Step 2.3.2.2

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

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

Step 2.3.2.3

Multiply the new quotient term by the divisor.

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

Step 2.3.2.4

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

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

Step 2.3.2.5

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

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

Step 2.3.2.6

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

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

Step 2.3.2.7

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

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

Step 2.3.2.8

Multiply the new quotient term by the divisor.

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

Step 2.3.2.9

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

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

Step 2.3.2.10

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

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

Step 2.3.2.11

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

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

Step 2.3.3

Split the single integral into multiple integrals.

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Apply the constant rule.

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

Step 2.3.7

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

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

Step 2.3.8

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

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Step 2.3.8.1

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

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Step 2.3.8.1.1

Rewrite.

$\frac{- 1}{1}$

Step 2.3.8.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.3.8.2

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

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

Step 2.3.9

Move the negative in front of the fraction.

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

Step 2.3.10

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

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

Step 2.3.11

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

$- \frac{1}{y} + C_{1} = - (\frac{x^{2}}{2} + C_{2}) - x + C_{3} - (\ln (| u |) + C_{4})$

Step 2.3.12

Simplify.

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

Step 2.3.13

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

To write $- \ln (| - x + 1 |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.1.3

Simplify terms.

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Step 3.1.3.1

Combine $- \ln (| - x + 1 |)$ and $\frac{2}{2}$ .

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

Step 3.1.3.2

Combine the numerators over the common denominator.

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

Step 3.1.4

Simplify the numerator.

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Step 3.1.4.1

Multiply $- \ln (| - x + 1 |) \cdot 2$ .

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Step 3.1.4.1.1

Multiply $2$ by $- 1$ .

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

Step 3.1.4.1.2

Simplify $- 2 \ln (| - x + 1 |)$ by moving $2$ inside the logarithm.

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

Step 3.1.4.2

Remove the absolute value in ${| - x + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 1, 2, 1$

Step 3.2.2

Since $y, 1, 2, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 2, 1$ then find LCM for the variable part $y^{1}$ .

Step 3.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 3.2.5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 3.2.6

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 3.2.7

The LCM of $1, 1, 2, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

Step 3.2.8

The factor for $y^{1}$ is $y$ itself.

$y^{1} = y$

$y$ occurs $1$ time.

Step 3.2.9

The LCM of $y^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$y$

Step 3.2.10

The LCM for $y, 1, 2, 1$ is the numeric part $2$ multiplied by the variable part.

$2 y$

Step 3.3

Multiply each term in $- \frac{1}{y} = - x + \frac{- x^{2} - \ln ({(- x + 1)}^{2})}{2} + C$ by $2 y$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $- \frac{1}{y} = - x + \frac{- x^{2} - \ln ({(- x + 1)}^{2})}{2} + C$ by $2 y$ .

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $y$ .

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Step 3.3.2.1.1

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

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

Step 3.3.2.1.2

Factor $y$ out of $2 y$ .

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

Step 3.3.2.1.3

Cancel the common factor.

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

Step 3.3.2.1.4

Rewrite the expression.

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

Step 3.3.2.2

Multiply $- 1$ by $2$ .

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Rewrite using the commutative property of multiplication.

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

Step 3.3.3.1.2

Multiply $- 1$ by $2$ .

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

Step 3.3.3.1.3

Rewrite using the commutative property of multiplication.

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

Step 3.3.3.1.4

Cancel the common factor of $2$ .

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Step 3.3.3.1.4.1

Cancel the common factor.

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

Step 3.3.3.1.4.2

Rewrite the expression.

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

Step 3.3.3.1.5

Apply the distributive property.

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

Step 3.3.3.1.6

Rewrite using the commutative property of multiplication.

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

Step 3.3.3.2

Reorder factors in $- 2 x y - x^{2} y - \ln ({(- x + 1)}^{2}) y + 2 C y$ .

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

Step 3.4

Solve the equation.

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Step 3.4.1

Rewrite the equation as $- 2 x y - x^{2} y - y \ln ({(- x + 1)}^{2}) + 2 C y = - 2$ .

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

Step 3.4.2

Factor $y$ out of $- 2 x y - x^{2} y - y \ln ({(- x + 1)}^{2}) + 2 C y$ .

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Step 3.4.2.1

Factor $y$ out of $- 2 x y$ .

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

Step 3.4.2.2

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

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

Step 3.4.2.3

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

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

Step 3.4.2.4

Factor $y$ out of $2 C y$ .

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

Step 3.4.2.5

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

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

Step 3.4.2.6

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

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

Step 3.4.2.7

Factor $y$ out of $y (- 2 x - x^{2} - 1 \ln ({(- x + 1)}^{2})) + y (2 C)$ .

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

Step 3.4.3

Rewrite $- 1 \ln ({(- x + 1)}^{2})$ as $- \ln ({(- x + 1)}^{2})$ .

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

Step 3.4.4

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

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Step 3.4.4.1

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

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Cancel the common factor of $- 2 x - x^{2} - \ln ({(- x + 1)}^{2}) + 2 C$ .

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Step 3.4.4.2.1.1

Cancel the common factor.

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

Step 3.4.4.2.1.2

Divide $y$ by $1$ .

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

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Move the negative in front of the fraction.

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

Step 3.4.4.3.2

Factor $- 1$ out of $- 2 x$ .

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

Step 3.4.4.3.3

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

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

Step 3.4.4.3.4

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

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

Step 3.4.4.3.5

Factor $- 1$ out of $- \ln ({(- x + 1)}^{2})$ .

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

Step 3.4.4.3.6

Factor $- 1$ out of $- (2 x + x^{2}) - (\ln ({(- x + 1)}^{2}))$ .

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

Step 3.4.4.3.7

Factor $- 1$ out of $2 C$ .

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

Step 3.4.4.3.8

Factor $- 1$ out of $- (2 x + x^{2} + \ln ({(- x + 1)}^{2})) - (- 2 C)$ .

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

Step 3.4.4.3.9

Simplify the expression.

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Step 3.4.4.3.9.1

Rewrite $- (2 x + x^{2} + \ln ({(- x + 1)}^{2}) - 2 C)$ as $- 1 (2 x + x^{2} + \ln ({(- x + 1)}^{2}) - 2 C)$ .

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

Step 3.4.4.3.9.2

Move the negative in front of the fraction.

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

Step 3.4.4.3.9.3

Multiply $- 1$ by $- 1$ .

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

Step 3.4.4.3.9.4

Multiply $\frac{2}{2 x + x^{2} + \ln ({(- x + 1)}^{2}) - 2 C}$ by $1$ .

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

Step 4

Simplify the constant of integration.

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