Calculus Examples

$(x^{3} y + 8 y) d x + (y + 1) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = x^{3} y + 8 y$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{3} y + 8 y]$

Step 1.2

By the Sum Rule, the derivative of $x^{3} y + 8 y$ with respect to $y$ is $\frac{d}{d y} [x^{3} y] + \frac{d}{d y} [8 y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [x^{3} y] + \frac{d}{d y} [8 y]$

Step 1.3

Evaluate $\frac{d}{d y} [x^{3} y]$ .

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = x^{3} \frac{d}{d y} [y] + \frac{d}{d y} [8 y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = x^{3} \cdot 1 + \frac{d}{d y} [8 y]$

Step 1.3.3

Multiply $x^{3}$ by $1$ .

$\frac{\partial M}{\partial y} = x^{3} + \frac{d}{d y} [8 y]$

Step 1.4

Evaluate $\frac{d}{d y} [8 y]$ .

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Step 1.4.1

Since $8$ is constant with respect to $y$ , the derivative of $8 y$ with respect to $y$ is $8 \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = x^{3} + 8 \frac{d}{d y} [y]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = x^{3} + 8 \cdot 1$

Step 1.4.3

Multiply $8$ by $1$ .

$\frac{\partial M}{\partial y} = x^{3} + 8$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = y + 1$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y + 1]$

Step 2.2

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

$\frac{\partial N}{\partial x} = \frac{d}{d x} [y] + \frac{d}{d x} [1]$

Step 2.3

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [1]$

Step 2.4

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

$\frac{\partial N}{\partial x} = 0 + 0$

Step 2.5

Add $0$ and $0$ .

$\frac{\partial N}{\partial x} = 0$

Step 3

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

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Step 3.1

Substitute $x^{3} + 8$ for $\frac{\partial M}{\partial y}$ and $0$ for $\frac{\partial N}{\partial x}$ .

$x^{3} + 8 = 0$

Step 3.2

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

$x^{3} + 8 = 0$ is not an identity.

Step 4

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

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Step 4.1

Substitute $0$ for $\frac{\partial N}{\partial x}$ .

$\frac{0 - \frac{\partial M}{\partial y}}{M}$

Step 4.2

Substitute $x^{3} + 8$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - (x^{3} + 8)}{M}$

Step 4.3

Substitute $x^{3} y + 8 y$ for $M$ .

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Step 4.3.1

Substitute $x^{3} y + 8 y$ for $M$ .

$\frac{0 - (x^{3} + 8)}{x^{3} y + 8 y}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{0 - x^{3} - 1 \cdot 8}{x^{3} y + 8 y}$

Step 4.3.2.2

Multiply $- 1$ by $8$ .

$\frac{0 - x^{3} - 8}{x^{3} y + 8 y}$

Step 4.3.2.3

Subtract $- (- x^{3} - 8)$ from $0$ .

$\frac{- x^{3} - 8}{x^{3} y + 8 y}$

Step 4.3.2.4

Rewrite $- x^{3} - 8$ in a factored form.

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Step 4.3.2.4.1

Rewrite $- x^{3}$ as ${(- x)}^{3}$ .

$\frac{{(- x)}^{3} - 8}{x^{3} y + 8 y}$

Step 4.3.2.4.2

Rewrite $8$ as $2^{3}$ .

$\frac{{(- x)}^{3} - 2^{3}}{x^{3} y + 8 y}$

Step 4.3.2.4.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = - x$ and $b = 2$ .

$\frac{(- x - 2) ({(- x)}^{2} - x \cdot 2 + 2^{2})}{x^{3} y + 8 y}$

Step 4.3.2.4.4

Simplify.

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Step 4.3.2.4.4.1

Apply the product rule to $- x$ .

$\frac{(- x - 2) ({(- 1)}^{2} x^{2} - x \cdot 2 + 2^{2})}{x^{3} y + 8 y}$

Step 4.3.2.4.4.2

Raise $- 1$ to the power of $2$ .

$\frac{(- x - 2) (1 x^{2} - x \cdot 2 + 2^{2})}{x^{3} y + 8 y}$

Step 4.3.2.4.4.3

Multiply $x^{2}$ by $1$ .

$\frac{(- x - 2) (x^{2} - x \cdot 2 + 2^{2})}{x^{3} y + 8 y}$

Step 4.3.2.4.4.4

Multiply $2$ by $- 1$ .

$\frac{(- x - 2) (x^{2} - 2 x + 2^{2})}{x^{3} y + 8 y}$

Step 4.3.2.4.4.5

Raise $2$ to the power of $2$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{x^{3} y + 8 y}$

Step 4.3.3

Simplify the denominator.

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Step 4.3.3.1

Factor $y$ out of $x^{3} y + 8 y$ .

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Step 4.3.3.1.1

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

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y x^{3} + 8 y}$

Step 4.3.3.1.2

Factor $y$ out of $8 y$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y x^{3} + y \cdot 8}$

Step 4.3.3.1.3

Factor $y$ out of $y x^{3} + y \cdot 8$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y (x^{3} + 8)}$

Step 4.3.3.2

Rewrite $8$ as $2^{3}$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y (x^{3} + 2^{3})}$

Step 4.3.3.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 2$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y ((x + 2) (x^{2} - x \cdot 2 + 2^{2}))}$

Step 4.3.3.4

Simplify.

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Step 4.3.3.4.1

Multiply $2$ by $- 1$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y ((x + 2) (x^{2} - 2 x + 2^{2}))}$

Step 4.3.3.4.2

Raise $2$ to the power of $2$ .

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y ((x + 2) (x^{2} - 2 x + 4))}$

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y (x + 2) (x^{2} - 2 x + 4)}$

Step 4.3.4

Cancel the common factor of $x^{2} - 2 x + 4$ .

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Step 4.3.4.1

Cancel the common factor.

$\frac{(- x - 2) (x^{2} - 2 x + 4)}{y (x + 2) (x^{2} - 2 x + 4)}$

Step 4.3.4.2

Rewrite the expression.

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

Step 4.3.5

Cancel the common factor of $- x - 2$ and $x + 2$ .

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Step 4.3.5.1

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

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

Step 4.3.5.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 4.3.5.3

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

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

Step 4.3.5.4

Rewrite $- (x + 2)$ as $- 1 (x + 2)$ .

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

Step 4.3.5.5

Cancel the common factor.

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

Step 4.3.5.6

Rewrite the expression.

$\frac{- 1}{y}$

Step 4.3.6

Substitute $x^{3} y + 8 y$ for $M$ .

$- \frac{1}{y}$

Step 4.4

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

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

Step 5

Evaluate the integral $e^{\int - \frac{1}{y} d y}$ .

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{- (\ln (| y |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{- \ln (| y |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

Simplify $- \ln (| y |)$ by moving $- 1$ inside the logarithm.

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 1} + C$

Step 5.4.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{| y |} + C$

Step 6

Multiply both sides of $(x^{3} y + 8 y) d x + (y + 1) d y = 0$ by the integration factor $\frac{1}{y}$ .

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Step 6.1

Multiply $x^{3} y + 8 y$ by $\frac{1}{y}$ .

$(x^{3} y + 8 y) \frac{1}{y} d x + (y + 1) d y = 0$

Step 6.2

Multiply $x^{3} y + 8 y$ by $\frac{1}{y}$ .

$\frac{x^{3} y + 8 y}{y} d x + (y + 1) d y = 0$

Step 6.3

Simplify the numerator.

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Step 6.3.1

Factor $y$ out of $x^{3} y + 8 y$ .

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Step 6.3.1.1

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

$\frac{y x^{3} + 8 y}{y} d x + (y + 1) d y = 0$

Step 6.3.1.2

Factor $y$ out of $8 y$ .

$\frac{y x^{3} + y \cdot 8}{y} d x + (y + 1) d y = 0$

Step 6.3.1.3

Factor $y$ out of $y x^{3} + y \cdot 8$ .

$\frac{y (x^{3} + 8)}{y} d x + (y + 1) d y = 0$

Step 6.3.2

Rewrite $8$ as $2^{3}$ .

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

Step 6.3.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 2$ .

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

Step 6.3.4

Simplify.

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Step 6.3.4.1

Multiply $2$ by $- 1$ .

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

Step 6.3.4.2

Raise $2$ to the power of $2$ .

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

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

Step 6.4

Cancel the common factor of $y$ .

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Step 6.4.1

Cancel the common factor.

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

Step 6.4.2

Divide $(x + 2) (x^{2} - 2 x + 4)$ by $1$ .

$(x + 2) (x^{2} - 2 x + 4) d x + (y + 1) d y = 0$

Step 6.5

Expand $(x + 2) (x^{2} - 2 x + 4)$ by multiplying each term in the first expression by each term in the second expression.

$(x \cdot x^{2} + x (- 2 x) + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6

Simplify each term.

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Step 6.6.1

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

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Step 6.6.1.1

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

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Step 6.6.1.1.1

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

$(x^{1} x^{2} + x (- 2 x) + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.1.1.2

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

$(x^{1 + 2} + x (- 2 x) + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.1.2

Add $1$ and $2$ .

$(x^{3} + x (- 2 x) + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.2

Rewrite using the commutative property of multiplication.

$(x^{3} - 2 x \cdot x + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.3

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

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Step 6.6.3.1

Move $x$ .

$(x^{3} - 2 (x \cdot x) + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.3.2

Multiply $x$ by $x$ .

$(x^{3} - 2 x^{2} + x \cdot 4 + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.4

Move $4$ to the left of $x$ .

$(x^{3} - 2 x^{2} + 4 \cdot x + 2 x^{2} + 2 (- 2 x) + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.5

Multiply $- 2$ by $2$ .

$(x^{3} - 2 x^{2} + 4 x + 2 x^{2} - 4 x + 2 \cdot 4) d x + (y + 1) d y = 0$

Step 6.6.6

Multiply $2$ by $4$ .

$(x^{3} - 2 x^{2} + 4 x + 2 x^{2} - 4 x + 8) d x + (y + 1) d y = 0$

Step 6.7

Combine the opposite terms in $x^{3} - 2 x^{2} + 4 x + 2 x^{2} - 4 x + 8$ .

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Step 6.7.1

Add $- 2 x^{2}$ and $2 x^{2}$ .

$(x^{3} + 4 x + 0 - 4 x + 8) d x + (y + 1) d y = 0$

Step 6.7.2

Add $x^{3} + 4 x$ and $0$ .

$(x^{3} + 4 x - 4 x + 8) d x + (y + 1) d y = 0$

Step 6.7.3

Subtract $4 x$ from $4 x$ .

$(x^{3} + 0 + 8) d x + (y + 1) d y = 0$

Step 6.7.4

Add $x^{3}$ and $0$ .

$(x^{3} + 8) d x + (y + 1) d y = 0$

Step 6.8

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

$(x^{3} + 8) d x + (y + 1) \frac{1}{y} d y = 0$

Step 6.9

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

$(x^{3} + 8) d x + \frac{y + 1}{y} d y = 0$

Step 7

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

$f (x, y) = \int x^{3} + 8 d x$

Step 8

Integrate $M (x, y) = x^{3} + 8$ to find $f (x, y)$ .

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Step 8.1

Split the single integral into multiple integrals.

$f (x, y) = \int x^{3} d x + \int 8 d x$

Step 8.2

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$f (x, y) = \frac{1}{4} x^{4} + C + \int 8 d x$

Step 8.3

Apply the constant rule.

$f (x, y) = \frac{1}{4} x^{4} + C + 8 x + C$

Step 8.4

Simplify.

$f (x, y) = \frac{1}{4} x^{4} + 8 x + C$

Step 9

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

$f (x, y) = \frac{1}{4} x^{4} + 8 x + g (y)$

Step 10

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

$\frac{\partial f}{\partial y} = \frac{y + 1}{y}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $y$ .

$\frac{d}{d y} [\frac{1}{4} x^{4} + 8 x + g (y)] = \frac{y + 1}{y}$

Step 11.2

By the Sum Rule, the derivative of $\frac{1}{4} x^{4} + 8 x + g (y)$ with respect to $y$ is $\frac{d}{d y} [\frac{1}{4} x^{4}] + \frac{d}{d y} [8 x] + \frac{d}{d y} [g (y)]$ .

$\frac{d}{d y} [\frac{1}{4} x^{4}] + \frac{d}{d y} [8 x] + \frac{d}{d y} [g (y)] = \frac{y + 1}{y}$

Step 11.3

Since $\frac{1}{4} x^{4}$ is constant with respect to $y$ , the derivative of $\frac{1}{4} x^{4}$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [8 x] + \frac{d}{d y} [g (y)] = \frac{y + 1}{y}$

Step 11.4

Since $8 x$ is constant with respect to $y$ , the derivative of $8 x$ with respect to $y$ is $0$ .

$0 + 0 + \frac{d}{d y} [g (y)] = \frac{y + 1}{y}$

Step 11.5

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

$0 + 0 + \frac{d g}{d y} = \frac{y + 1}{y}$

Step 11.6

Combine terms.

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Step 11.6.1

Add $0$ and $0$ .

$0 + \frac{d g}{d y} = \frac{y + 1}{y}$

Step 11.6.2

Add $0$ and $\frac{d g}{d y}$ .

$\frac{d g}{d y} = \frac{y + 1}{y}$

Step 12

Find the antiderivative of $\frac{y + 1}{y}$ to find $g (y)$ .

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Step 12.1

Integrate both sides of $\frac{d g}{d y} = \frac{y + 1}{y}$ .

$\int \frac{d g}{d y} d y = \int \frac{y + 1}{y} d y$

Step 12.2

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

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

Step 12.3

Split the fraction into multiple fractions.

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

Step 12.4

Split the single integral into multiple integrals.

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

Step 12.5

Cancel the common factor of $y$ .

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Step 12.5.1

Cancel the common factor.

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

Step 12.5.2

Rewrite the expression.

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

Step 12.6

Apply the constant rule.

$g (y) = y + C + \int \frac{1}{y} d y$

Step 12.7

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

$g (y) = y + C + \ln (| y |) + C$

Step 12.8

Simplify.

$g (y) = y + \ln (| y |) + C$

Step 13

Substitute for $g (y)$ in $f (x, y) = \frac{1}{4} x^{4} + 8 x + g (y)$ .

$f (x, y) = \frac{1}{4} x^{4} + 8 x + y + \ln (| y |) + C$

Step 14

Combine $\frac{1}{4}$ and $x^{4}$ .

$f (x, y) = \frac{x^{4}}{4} + 8 x + y + \ln (| y |) + C$