Calculus Examples

$(2 x y - y) d x + (x^{2} + x) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x y - y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y - y]$

Step 1.2

By the Sum Rule, the derivative of $2 x y - y$ with respect to $y$ is $\frac{d}{d y} [2 x y] + \frac{d}{d y} [- y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x y] + \frac{d}{d y} [- y]$

Step 1.3

Evaluate $\frac{d}{d y} [2 x y]$ .

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Step 1.3.1

Since $2 x$ is constant with respect to $y$ , the derivative of $2 x y$ with respect to $y$ is $2 x \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = 2 x \frac{d}{d y} [y] + \frac{d}{d y} [- y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 2 x \cdot 1 + \frac{d}{d y} [- y]$

Step 1.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x + \frac{d}{d y} [- y]$

Step 1.4

Evaluate $\frac{d}{d y} [- y]$ .

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Step 1.4.1

Since $- 1$ is constant with respect to $y$ , the derivative of $- y$ with respect to $y$ is $- \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = 2 x - \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 2 x - 1 \cdot 1$

Step 1.4.3

Multiply $- 1$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x - 1$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x^{2} + x$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{2} + x]$

Step 2.2

By the Sum Rule, the derivative of $x^{2} + x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x]$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{\partial N}{\partial x} = 2 x + \frac{d}{d x} [x]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 2 x + 1$

Step 3

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 3.1

Substitute $2 x - 1$ for $\frac{\partial M}{\partial y}$ and $2 x + 1$ for $\frac{\partial N}{\partial x}$ .

$2 x - 1 = 2 x + 1$

Step 3.2

Since the left side does not equal the right side, the equation is not an identity.

$2 x - 1 = 2 x + 1$ is not an identity.

Step 4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

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Step 4.1

Substitute $2 x - 1$ for $\frac{\partial M}{\partial y}$ .

$\frac{2 x - 1 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $2 x + 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{2 x - 1 - (2 x + 1)}{N}$

Step 4.3

Substitute $x^{2} + x$ for $N$ .

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Step 4.3.1

Substitute $x^{2} + x$ for $N$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{2 x - 1 - (2 x) - 1 \cdot 1}{x^{2} + x}$

Step 4.3.2.2

Multiply $2$ by $- 1$ .

$\frac{2 x - 1 - 2 x - 1 \cdot 1}{x^{2} + x}$

Step 4.3.2.3

Multiply $- 1$ by $1$ .

$\frac{2 x - 1 - 2 x - 1}{x^{2} + x}$

Step 4.3.2.4

Subtract $2 x$ from $2 x$ .

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

Step 4.3.2.5

Subtract $1$ from $0$ .

$\frac{- 1 - 1}{x^{2} + x}$

Step 4.3.2.6

Subtract $1$ from $- 1$ .

$\frac{- 2}{x^{2} + x}$

Step 4.3.3

Factor $x$ out of $x^{2} + x$ .

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Step 4.3.3.1

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

$\frac{- 2}{x \cdot x + x}$

Step 4.3.3.2

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

$\frac{- 2}{x \cdot x + x^{1}}$

Step 4.3.3.3

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

$\frac{- 2}{x \cdot x + x \cdot 1}$

Step 4.3.3.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$\frac{- 2}{x (x + 1)}$

Step 4.3.4

Move the negative in front of the fraction.

$- \frac{2}{x (x + 1)}$

Step 4.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

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

Step 5

Evaluate the integral $e^{\int - \frac{2}{x (x + 1)} d x}$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

Write the fraction using partial fraction decomposition.

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Step 5.4.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.4.1.1

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{x} + \frac{B}{x + 1}$

Step 5.4.1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x + 1)$ .

$\frac{1 (x (x + 1))}{x (x + 1)} = \frac{A (x (x + 1))}{x} + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.3

Cancel the common factor of $x$ .

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Step 5.4.1.3.1

Cancel the common factor.

$\frac{1 (x (x + 1))}{x (x + 1)} = \frac{A (x (x + 1))}{x} + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.3.2

Rewrite the expression.

$\frac{1 (x + 1)}{x + 1} = \frac{A (x (x + 1))}{x} + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.4

Cancel the common factor of $x + 1$ .

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Step 5.4.1.4.1

Cancel the common factor.

$\frac{1 (x + 1)}{x + 1} = \frac{A (x (x + 1))}{x} + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.4.2

Rewrite the expression.

$1 = \frac{A (x (x + 1))}{x} + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.5

Simplify each term.

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Step 5.4.1.5.1

Cancel the common factor of $x$ .

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Step 5.4.1.5.1.1

Cancel the common factor.

$1 = \frac{A (x (x + 1))}{x} + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.5.1.2

Divide $A (x + 1)$ by $1$ .

$1 = A (x + 1) + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.5.2

Apply the distributive property.

$1 = A x + A \cdot 1 + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.5.3

Multiply $A$ by $1$ .

$1 = A x + A + \frac{(B) (x (x + 1))}{x + 1}$

Step 5.4.1.5.4

Cancel the common factor of $x + 1$ .

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Step 5.4.1.5.4.1

Cancel the common factor.

$1 = A x + A + \frac{B (x (x + 1))}{x + 1}$

Step 5.4.1.5.4.2

Divide $(B) (x)$ by $1$ .

$1 = A x + A + (B) (x)$

$1 = A x + A + B x$

Step 5.4.1.6

Move $A$ .

$1 = A x + B x + A$

Step 5.4.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 5.4.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 5.4.2.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A$

Step 5.4.2.3

Set up the system of equations to find the coefficients of the partial fractions.

$0 = A + B$

$1 = A$

$0 = A + B$

$1 = A$

Step 5.4.3

Solve the system of equations.

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Step 5.4.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + B$

Step 5.4.3.2

Replace all occurrences of $A$ with $1$ in each equation.

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Step 5.4.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $1$ .

$0 = (1) + B$

$A = 1$

Step 5.4.3.2.2

Simplify the right side.

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Step 5.4.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

Step 5.4.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 5.4.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

Step 5.4.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

Step 5.4.3.4

Solve the system of equations.

$B = - 1 A = 1$

Step 5.4.3.5

List all of the solutions.

$B = - 1, A = 1$

Step 5.4.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x + 1}$ with the values found for $A$ and $B$ .

$\frac{1}{x} + \frac{- 1}{x + 1}$

Step 5.4.5

Move the negative in front of the fraction.

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

Step 5.5

Split the single integral into multiple integrals.

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

Step 5.6

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

$μ (x, y) = e^{- 2 (\ln (| x |) + C + \int - \frac{1}{x + 1} d x)}$

Step 5.7

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

$μ (x, y) = e^{- 2 (\ln (| x |) + C - \int \frac{1}{x + 1} d x)}$

Step 5.8

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

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Step 5.8.1

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

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Step 5.8.1.1

Differentiate $x + 1$ .

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

Step 5.8.1.2

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 5.8.1.3

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 5.8.1.4

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

$1 + 0$

Step 5.8.1.5

Add $1$ and $0$ .

$1$

Step 5.8.2

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

$μ (x, y) = e^{- 2 (\ln (| x |) + C - \int \frac{1}{u} d u)}$

Step 5.9

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

$μ (x, y) = e^{- 2 (\ln (| x |) + C - (\ln (| u |) + C))}$

Step 5.10

Simplify.

$μ (x, y) = e^{- 2 \ln (\frac{| x |}{| u |})} + C$

Step 5.11

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

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

Step 5.12

Simplify each term.

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Step 5.12.1

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

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

Step 5.12.2

Exponentiation and log are inverse functions.

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

Step 5.12.3

Change the sign of the exponent by rewriting the base as its reciprocal.

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

Step 5.12.4

Apply the product rule to $\frac{| x + 1 |}{| x |}$ .

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

Step 5.12.5

Simplify the numerator.

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Step 5.12.5.1

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

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

Step 5.12.5.2

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

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

Step 5.12.5.3

Expand $(x + 1) (x + 1)$ using the FOIL Method.

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Step 5.12.5.3.1

Apply the distributive property.

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

Step 5.12.5.3.2

Apply the distributive property.

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

Step 5.12.5.3.3

Apply the distributive property.

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

Step 5.12.5.4

Simplify and combine like terms.

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Step 5.12.5.4.1

Simplify each term.

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Step 5.12.5.4.1.1

Multiply $x$ by $x$ .

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

Step 5.12.5.4.1.2

Multiply $x$ by $1$ .

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

Step 5.12.5.4.1.3

Multiply $x$ by $1$ .

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

Step 5.12.5.4.1.4

Multiply $1$ by $1$ .

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

Step 5.12.5.4.2

Add $x$ and $x$ .

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

Step 5.12.5.5

Factor using the perfect square rule.

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Step 5.12.5.5.1

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

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

Step 5.12.5.5.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 5.12.5.5.3

Rewrite the polynomial.

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

Step 5.12.5.5.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 5.12.6

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

$μ (x, y) = \frac{{(x + 1)}^{2}}{x^{2}} + C$

Step 6

Multiply both sides of $(2 x y - y) d x + (x^{2} + x) d y = 0$ by the integration factor $\frac{{(x + 1)}^{2}}{x^{2}}$ .

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Step 6.1

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

$(2 x y - y) \frac{{(x + 1)}^{2}}{x^{2}} d x + (x^{2} + x) d y = 0$

Step 6.2

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

$\frac{(2 x y - y) {(x + 1)}^{2}}{x^{2}} d x + (x^{2} + x) d y = 0$

Step 6.3

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

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Step 6.3.1

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

$\frac{(y (2 x) - y) {(x + 1)}^{2}}{x^{2}} d x + (x^{2} + x) d y = 0$

Step 6.3.2

Factor $y$ out of $- y$ .

$\frac{(y (2 x) + y \cdot - 1) {(x + 1)}^{2}}{x^{2}} d x + (x^{2} + x) d y = 0$

Step 6.3.3

Factor $y$ out of $y (2 x) + y \cdot - 1$ .

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + (x^{2} + x) d y = 0$

Step 6.4

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + (x^{2} + x) \frac{{(x + 1)}^{2}}{x^{2}} d y = 0$

Step 6.5

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{(x^{2} + x) {(x + 1)}^{2}}{x^{2}} d y = 0$

Step 6.6

Simplify the numerator.

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Step 6.6.1

Factor $x$ out of $x^{2} + x$ .

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Step 6.6.1.1

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{(x \cdot x + x) {(x + 1)}^{2}}{x^{2}} d y = 0$

Step 6.6.1.2

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{(x \cdot x + x^{1}) {(x + 1)}^{2}}{x^{2}} d y = 0$

Step 6.6.1.3

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{(x \cdot x + x \cdot 1) {(x + 1)}^{2}}{x^{2}} d y = 0$

Step 6.6.1.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x (x + 1) {(x + 1)}^{2}}{x^{2}} d y = 0$

Step 6.6.2

Multiply $x + 1$ by ${(x + 1)}^{2}$ by adding the exponents.

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Step 6.6.2.1

Move ${(x + 1)}^{2}$ .

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x ({(x + 1)}^{2} (x + 1))}{x^{2}} d y = 0$

Step 6.6.2.2

Multiply ${(x + 1)}^{2}$ by $x + 1$ .

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Step 6.6.2.2.1

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x ({(x + 1)}^{2} {(x + 1)}^{1})}{x^{2}} d y = 0$

Step 6.6.2.2.2

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x {(x + 1)}^{2 + 1}}{x^{2}} d y = 0$

Step 6.6.2.3

Add $2$ and $1$ .

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x {(x + 1)}^{3}}{x^{2}} d y = 0$

Step 6.7

Cancel the common factor of $x$ and $x^{2}$ .

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Step 6.7.1

Factor $x$ out of $x {(x + 1)}^{3}$ .

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x ({(x + 1)}^{3})}{x^{2}} d y = 0$

Step 6.7.2

Cancel the common factors.

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Step 6.7.2.1

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

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x ({(x + 1)}^{3})}{x \cdot x} d y = 0$

Step 6.7.2.2

Cancel the common factor.

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{x {(x + 1)}^{3}}{x \cdot x} d y = 0$

Step 6.7.2.3

Rewrite the expression.

$\frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}} d x + \frac{{(x + 1)}^{3}}{x} d y = 0$

Step 7

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int \frac{{(x + 1)}^{3}}{x} d y$

Step 8

Integrate $N (x, y) = \frac{{(x + 1)}^{3}}{x}$ to find $f (x, y)$ .

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Step 8.1

Apply the constant rule.

$f (x, y) = \frac{{(x + 1)}^{3}}{x} y + C$

Step 8.2

Simplify the answer.

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Step 8.2.1

Combine $\frac{{(x + 1)}^{3}}{x}$ and $y$ .

$f (x, y) = \frac{{(x + 1)}^{3} y}{x} + C$

Step 8.2.2

Rewrite $\frac{{(x + 1)}^{3} y}{x} + C$ as $\frac{(x^{3} + 3 x^{2} \cdot 1 + 3 x \cdot 1^{2} + 1^{3}) y}{x} + C$ .

$f (x, y) = \frac{(x^{3} + 3 x^{2} \cdot 1 + 3 x \cdot 1^{2} + 1^{3}) y}{x} + C$

Step 8.2.3

Simplify.

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Step 8.2.3.1

Multiply $3$ by $1$ .

$f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x \cdot 1^{2} + 1^{3}) y}{x} + C$

Step 8.2.3.2

One to any power is one.

$f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x \cdot 1 + 1^{3}) y}{x} + C$

Step 8.2.3.3

Multiply $3$ by $1$ .

$f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x + 1^{3}) y}{x} + C$

Step 8.2.3.4

One to any power is one.

$f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + C$

Step 9

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + g (x)$

Step 10

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

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

Step 11

Find $\frac{\partial f}{\partial x}$ .

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Step 11.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [\frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.2

By the Sum Rule, the derivative of $\frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + g (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x}] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [\frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x}] + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3

Evaluate $\frac{d}{d x} [\frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x}]$ .

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Step 11.3.1

Since $y$ is constant with respect to $x$ , the derivative of $\frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x}$ with respect to $x$ is $y \frac{d}{d x} [\frac{x^{3} + 3 x^{2} + 3 x + 1}{x}]$ .

$y \frac{d}{d x} [\frac{x^{3} + 3 x^{2} + 3 x + 1}{x}] + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{3} + 3 x^{2} + 3 x + 1$ and $g (x) = x$ .

$y \frac{x \frac{d}{d x} [x^{3} + 3 x^{2} + 3 x + 1] - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.3

By the Sum Rule, the derivative of $x^{3} + 3 x^{2} + 3 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

$y \frac{x (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$y \frac{x (3 x^{2} + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.5

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$y \frac{x (3 x^{2} + 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$y \frac{x (3 x^{2} + 3 (2 x) + \frac{d}{d x} [3 x] + \frac{d}{d x} [1]) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.7

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$y \frac{x (3 x^{2} + 3 (2 x) + 3 \frac{d}{d x} [x] + \frac{d}{d x} [1]) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.8

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

$y \frac{x (3 x^{2} + 3 (2 x) + 3 \cdot 1 + \frac{d}{d x} [1]) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.9

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

$y \frac{x (3 x^{2} + 3 (2 x) + 3 \cdot 1 + 0) - (x^{3} + 3 x^{2} + 3 x + 1) \frac{d}{d x} [x]}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.10

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

$y \frac{x (3 x^{2} + 3 (2 x) + 3 \cdot 1 + 0) - (x^{3} + 3 x^{2} + 3 x + 1) \cdot 1}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.11

Multiply $2$ by $3$ .

$y \frac{x (3 x^{2} + 6 x + 3 \cdot 1 + 0) - (x^{3} + 3 x^{2} + 3 x + 1) \cdot 1}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.12

Multiply $3$ by $1$ .

$y \frac{x (3 x^{2} + 6 x + 3 + 0) - (x^{3} + 3 x^{2} + 3 x + 1) \cdot 1}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.13

Add $3 x^{2} + 6 x + 3$ and $0$ .

$y \frac{x (3 x^{2} + 6 x + 3) - (x^{3} + 3 x^{2} + 3 x + 1) \cdot 1}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.14

Multiply $- 1$ by $1$ .

$y \frac{x (3 x^{2} + 6 x + 3) - (x^{3} + 3 x^{2} + 3 x + 1)}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.3.15

Combine $y$ and $\frac{x (3 x^{2} + 6 x + 3) - (x^{3} + 3 x^{2} + 3 x + 1)}{x^{2}}$ .

$\frac{y (x (3 x^{2} + 6 x + 3) - (x^{3} + 3 x^{2} + 3 x + 1))}{x^{2}} + \frac{d}{d x} [g (x)] = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.4

Differentiate using the function rule which states that the derivative of $g (x)$ is $\frac{d g}{d x}$ .

$\frac{y (x (3 x^{2} + 6 x + 3) - (x^{3} + 3 x^{2} + 3 x + 1))}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5

Simplify.

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Step 11.5.1

Apply the distributive property.

$\frac{y (x (3 x^{2}) + x (6 x) + x \cdot 3 - (x^{3} + 3 x^{2} + 3 x + 1))}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.2

Apply the distributive property.

$\frac{y (x (3 x^{2}) + x (6 x) + x \cdot 3 - x^{3} - (3 x^{2}) - (3 x) - 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.3

Apply the distributive property.

$\frac{y (x (3 x^{2})) + y (x (6 x)) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4

Combine terms.

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Step 11.5.4.1

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

$\frac{y (3 (x^{1} x^{2})) + y (x (6 x)) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.2

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

$\frac{y (3 x^{1 + 2}) + y (x (6 x)) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.3

Add $1$ and $2$ .

$\frac{y (3 x^{3}) + y (x (6 x)) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.4

Move $3$ to the left of $y$ .

$\frac{3 \cdot y x^{3} + y (x (6 x)) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.5

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

$\frac{3 y x^{3} + y (6 (x^{1} x)) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.6

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

$\frac{3 y x^{3} + y (6 (x^{1} x^{1})) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.7

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

$\frac{3 y x^{3} + y (6 x^{1 + 1}) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.8

Add $1$ and $1$ .

$\frac{3 y x^{3} + y (6 x^{2}) + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.9

Move $6$ to the left of $y$ .

$\frac{3 y x^{3} + 6 \cdot y x^{2} + y (x \cdot 3) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.10

Move $3$ to the left of $x$ .

$\frac{3 y x^{3} + 6 y x^{2} + y (3 \cdot x) + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.11

Move $3$ to the left of $y$ .

$\frac{3 y x^{3} + 6 y x^{2} + 3 \cdot y x + y (- x^{3}) + y (- (3 x^{2})) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.12

Multiply $3$ by $- 1$ .

$\frac{3 y x^{3} + 6 y x^{2} + 3 y x + y (- x^{3}) + y (- 3 x^{2}) + y (- (3 x)) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.13

Multiply $3$ by $- 1$ .

$\frac{3 y x^{3} + 6 y x^{2} + 3 y x + y (- x^{3}) + y (- 3 x^{2}) + y (- 3 x) + y (- 1 \cdot 1)}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.14

Multiply $- 1$ by $1$ .

$\frac{3 y x^{3} + 6 y x^{2} + 3 y x + y (- x^{3}) + y (- 3 x^{2}) + y (- 3 x) + y \cdot - 1}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.15

Move $- 1$ to the left of $y$ .

$\frac{3 y x^{3} + 6 y x^{2} + 3 y x + y (- x^{3}) + y (- 3 x^{2}) + y (- 3 x) - 1 \cdot y}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.16

Rewrite $- 1 y$ as $- y$ .

$\frac{3 y x^{3} + 6 y x^{2} + 3 y x + y (- x^{3}) + y (- 3 x^{2}) + y (- 3 x) - y}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.17

Move $y$ .

$\frac{3 x^{3} y - x^{3} y + 6 y x^{2} + 3 y x + y (- 3 x^{2}) + y (- 3 x) - y}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.18

Subtract $x^{3} y$ from $3 x^{3} y$ .

$\frac{2 x^{3} y + 6 y x^{2} + 3 y x + y (- 3 x^{2}) + y (- 3 x) - y}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.19

Move $y$ .

$\frac{2 x^{3} y + 6 x^{2} y - 3 x^{2} y + 3 y x + y (- 3 x) - y}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.20

Subtract $3 x^{2} y$ from $6 x^{2} y$ .

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

Step 11.5.4.21

Move $y$ .

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

Step 11.5.4.22

Subtract $3 x y$ from $3 x y$ .

$\frac{2 x^{3} y + 3 x^{2} y + 0 - y}{x^{2}} + \frac{d g}{d x} = \frac{y (2 x - 1) {(x + 1)}^{2}}{x^{2}}$

Step 11.5.4.23

Add $2 x^{3} y + 3 x^{2} y$ and $0$ .

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

Step 11.5.5

Reorder terms.

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

Step 11.5.6

Factor $y$ out of $2 x^{3} y + 3 x^{2} y - y$ .

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Step 11.5.6.1

Factor $y$ out of $2 x^{3} y$ .

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

Step 11.5.6.2

Factor $y$ out of $3 x^{2} y$ .

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

Step 11.5.6.3

Factor $y$ out of $- y$ .

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

Step 11.5.6.4

Factor $y$ out of $y (2 x^{3}) + y (3 x^{2})$ .

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

Step 11.5.6.5

Factor $y$ out of $y (2 x^{3} + 3 x^{2}) + y \cdot - 1$ .

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

Step 11.5.7

To write $\frac{d g}{d x}$ as a fraction with a common denominator, multiply by $\frac{x^{2}}{x^{2}}$ .

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

Step 11.5.8

Combine the numerators over the common denominator.

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

Step 11.5.9

Simplify the numerator.

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Step 11.5.9.1

Apply the distributive property.

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

Step 11.5.9.2

Simplify.

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Step 11.5.9.2.1

Rewrite using the commutative property of multiplication.

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

Step 11.5.9.2.2

Rewrite using the commutative property of multiplication.

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

Step 11.5.9.2.3

Move $- 1$ to the left of $y$ .

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

Step 11.5.9.3

Simplify each term.

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

Step 11.5.10

Reorder factors in $\frac{\frac{d g}{d x} x^{2} + 2 y x^{3} + 3 y x^{2} - y}{x^{2}}$ .

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

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

Step 12.1.2

Simplify $y (2 x - 1) {(x + 1)}^{2}$ .

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Step 12.1.2.1

Rewrite.

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 0 + 0 + y (2 x - 1) {(x + 1)}^{2}$

Step 12.1.2.2

Simplify terms.

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Step 12.1.2.2.1

Add $0$ and $0$ .

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 0 + y (2 x - 1) {(x + 1)}^{2}$

Step 12.1.2.2.2

Simplify by multiplying through.

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Step 12.1.2.2.2.1

Apply the distributive property.

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

Step 12.1.2.2.2.2

Reorder.

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Step 12.1.2.2.2.2.1

Rewrite using the commutative property of multiplication.

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

Step 12.1.2.2.2.2.2

Move $- 1$ to the left of $y$ .

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

Step 12.1.2.2.3

Rewrite $- 1 y$ as $- y$ .

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

Step 12.1.2.2.4

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

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

Step 12.1.2.3

Expand $(x + 1) (x + 1)$ using the FOIL Method.

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Step 12.1.2.3.1

Apply the distributive property.

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

Step 12.1.2.3.2

Apply the distributive property.

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

Step 12.1.2.3.3

Apply the distributive property.

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

Step 12.1.2.4

Simplify and combine like terms.

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Step 12.1.2.4.1

Simplify each term.

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Step 12.1.2.4.1.1

Multiply $x$ by $x$ .

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

Step 12.1.2.4.1.2

Multiply $x$ by $1$ .

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

Step 12.1.2.4.1.3

Multiply $x$ by $1$ .

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

Step 12.1.2.4.1.4

Multiply $1$ by $1$ .

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

Step 12.1.2.4.2

Add $x$ and $x$ .

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

Step 12.1.2.5

Expand $(2 y x - y) (x^{2} + 2 x + 1)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 12.1.2.6

Simplify terms.

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Step 12.1.2.6.1

Combine the opposite terms in $2 y x \cdot x^{2} + 2 y x (2 x) + 2 y x \cdot 1 - y x^{2} - y (2 x) - y \cdot 1$ .

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Step 12.1.2.6.1.1

Reorder the factors in the terms $2 y x \cdot 1$ and $- y (2 x)$ .

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

Step 12.1.2.6.1.2

Subtract $2 x y$ from $1 \cdot 2 x y$ .

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 2 y x \cdot x^{2} + 2 y x (2 x) - y x^{2} + 0 - y \cdot 1$

Step 12.1.2.6.1.3

Add $2 y x \cdot x^{2} + 2 y x (2 x) - y x^{2}$ and $0$ .

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

Step 12.1.2.6.2

Simplify each term.

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Step 12.1.2.6.2.1

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 12.1.2.6.2.1.1

Move $x^{2}$ .

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

Step 12.1.2.6.2.1.2

Multiply $x^{2}$ by $x$ .

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Step 12.1.2.6.2.1.2.1

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

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

Step 12.1.2.6.2.1.2.2

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

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

Step 12.1.2.6.2.1.3

Add $2$ and $1$ .

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

Step 12.1.2.6.2.2

Multiply $x$ by $x$ by adding the exponents.

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Step 12.1.2.6.2.2.1

Move $x$ .

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

Step 12.1.2.6.2.2.2

Multiply $x$ by $x$ .

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 2 y x^{3} + 2 y x^{2} \cdot 2 - y x^{2} - y \cdot 1$

Step 12.1.2.6.2.3

Multiply $2$ by $2$ .

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 2 y x^{3} + 4 y x^{2} - y x^{2} - y \cdot 1$

Step 12.1.2.6.2.4

Multiply $- 1$ by $1$ .

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 2 y x^{3} + 4 y x^{2} - y x^{2} - y$

Step 12.1.2.6.3

Subtract $y x^{2}$ from $4 y x^{2}$ .

$x^{2} \frac{d g}{d x} + 2 y x^{3} + 3 y x^{2} - y = 2 y x^{3} + 3 y x^{2} - y$

Step 12.1.3

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

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Step 12.1.3.1

Subtract $2 y x^{3}$ from both sides of the equation.

$x^{2} \frac{d g}{d x} + 3 y x^{2} - y = 2 y x^{3} + 3 y x^{2} - y - 2 y x^{3}$

Step 12.1.3.2

Subtract $3 y x^{2}$ from both sides of the equation.

$x^{2} \frac{d g}{d x} - y = 2 y x^{3} + 3 y x^{2} - y - 2 y x^{3} - 3 y x^{2}$

Step 12.1.3.3

Add $y$ to both sides of the equation.

$x^{2} \frac{d g}{d x} = 2 y x^{3} + 3 y x^{2} - y - 2 y x^{3} - 3 y x^{2} + y$

Step 12.1.3.4

Combine the opposite terms in $2 y x^{3} + 3 y x^{2} - y - 2 y x^{3} - 3 y x^{2} + y$ .

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Step 12.1.3.4.1

Subtract $2 y x^{3}$ from $2 y x^{3}$ .

$x^{2} \frac{d g}{d x} = 3 y x^{2} - y + 0 - 3 y x^{2} + y$

Step 12.1.3.4.2

Add $3 y x^{2} - y$ and $0$ .

$x^{2} \frac{d g}{d x} = 3 y x^{2} - y - 3 y x^{2} + y$

Step 12.1.3.4.3

Subtract $3 y x^{2}$ from $3 y x^{2}$ .

$x^{2} \frac{d g}{d x} = - y + 0 + y$

Step 12.1.3.4.4

Add $- y$ and $0$ .

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

Step 12.1.3.4.5

Add $- y$ and $y$ .

$x^{2} \frac{d g}{d x} = 0$

Step 12.1.4

Divide each term in $x^{2} \frac{d g}{d x} = 0$ by $x^{2}$ and simplify.

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Step 12.1.4.1

Divide each term in $x^{2} \frac{d g}{d x} = 0$ by $x^{2}$ .

$\frac{x^{2} \frac{d g}{d x}}{x^{2}} = \frac{0}{x^{2}}$

Step 12.1.4.2

Simplify the left side.

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Step 12.1.4.2.1

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

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Step 12.1.4.2.1.1

Cancel the common factor.

$\frac{x^{2} \frac{d g}{d x}}{x^{2}} = \frac{0}{x^{2}}$

Step 12.1.4.2.1.2

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

$\frac{d g}{d x} = \frac{0}{x^{2}}$

Step 12.1.4.3

Simplify the right side.

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Step 12.1.4.3.1

Divide $0$ by $x^{2}$ .

$\frac{d g}{d x} = 0$

Step 13

Find the antiderivative of $0$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = 0$ .

$\int \frac{d g}{d x} d x = \int 0 d x$

Step 13.2

Evaluate $\int \frac{d g}{d x} d x$ .

$g (x) = \int 0 d x$

Step 13.3

The integral of $0$ with respect to $x$ is $0$ .

$g (x) = 0 + C$

Step 13.4

Add $0$ and $C$ .

$g (x) = C$

Step 14

Substitute for $g (x)$ in $f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + g (x)$ .

$f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + C$

Step 15

Reorder factors in $f (x, y) = \frac{(x^{3} + 3 x^{2} + 3 x + 1) y}{x} + C$ .

$f (x, y) = \frac{y (x^{3} + 3 x^{2} + 3 x + 1)}{x} + C$