Calculus Examples

$(2 x^{2} y + 2 y + 5) d x + (2 x^{3} + 2 x) d y = 0$

Step 1

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 x^{2} y + 2 y + 5$ .

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x^{2} y + 2 y + 5]$

Step 1.2

By the Sum Rule, the derivative of $2 x^{2} y + 2 y + 5$ with respect to $y$ is $\frac{d}{d y} [2 x^{2} y] + \frac{d}{d y} [2 y] + \frac{d}{d y} [5]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 x^{2} y] + \frac{d}{d y} [2 y] + \frac{d}{d y} [5]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 2 x^{2} \frac{d}{d y} [y] + \frac{d}{d y} [2 y] + \frac{d}{d y} [5]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 2 x^{2} \cdot 1 + \frac{d}{d y} [2 y] + \frac{d}{d y} [5]$

Step 1.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x^{2} + \frac{d}{d y} [2 y] + \frac{d}{d y} [5]$

Step 1.4

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

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Step 1.4.1

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

$\frac{\partial M}{\partial y} = 2 x^{2} + 2 \frac{d}{d y} [y] + \frac{d}{d y} [5]$

Step 1.4.2

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

$\frac{\partial M}{\partial y} = 2 x^{2} + 2 \cdot 1 + \frac{d}{d y} [5]$

Step 1.4.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 x^{2} + 2 + \frac{d}{d y} [5]$

Step 1.5

Differentiate using the Constant Rule.

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Step 1.5.1

Since $5$ is constant with respect to $y$ , the derivative of $5$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 2 x^{2} + 2 + 0$

Step 1.5.2

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

$\frac{\partial M}{\partial y} = 2 x^{2} + 2$

Step 2

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = 2 x^{3} + 2 x$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 x^{3} + 2 x]$

Step 2.2

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

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [2 x]$

Step 2.3

Evaluate $\frac{d}{d x} [2 x^{3}]$ .

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [2 x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 2 (3 x^{2}) + \frac{d}{d x} [2 x]$

Step 2.3.3

Multiply $3$ by $2$ .

$\frac{\partial N}{\partial x} = 6 x^{2} + \frac{d}{d x} [2 x]$

Step 2.4

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

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 6 x^{2} + 2 \frac{d}{d x} [x]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 6 x^{2} + 2 \cdot 1$

Step 2.4.3

Multiply $2$ by $1$ .

$\frac{\partial N}{\partial x} = 6 x^{2} + 2$

Step 3

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

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Step 3.1

Substitute $2 x^{2} + 2$ for $\frac{\partial M}{\partial y}$ and $6 x^{2} + 2$ for $\frac{\partial N}{\partial x}$ .

$2 x^{2} + 2 = 6 x^{2} + 2$

Step 3.2

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

$2 x^{2} + 2 = 6 x^{2} + 2$ 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 $2 x^{2} + 2$ for $\frac{\partial M}{\partial y}$ .

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

Step 4.2

Substitute $6 x^{2} + 2$ for $\frac{\partial N}{\partial x}$ .

$\frac{2 x^{2} + 2 - (6 x^{2} + 2)}{N}$

Step 4.3

Substitute $2 x^{3} + 2 x$ for $N$ .

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Step 4.3.1

Substitute $2 x^{3} + 2 x$ for $N$ .

$\frac{2 x^{2} + 2 - (6 x^{2} + 2)}{2 x^{3} + 2 x}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

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

Step 4.3.2.2

Multiply $6$ by $- 1$ .

$\frac{2 x^{2} + 2 - 6 x^{2} - 1 \cdot 2}{2 x^{3} + 2 x}$

Step 4.3.2.3

Multiply $- 1$ by $2$ .

$\frac{2 x^{2} + 2 - 6 x^{2} - 2}{2 x^{3} + 2 x}$

Step 4.3.2.4

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

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

Step 4.3.2.5

Subtract $2$ from $2$ .

$\frac{- 4 x^{2} + 0}{2 x^{3} + 2 x}$

Step 4.3.2.6

Add $- 4 x^{2}$ and $0$ .

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

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

$\frac{- 4 x^{2}}{2 x (x^{2}) + 2 x (1)}$

Step 4.3.3.3

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

$\frac{- 4 x^{2}}{2 x (x^{2} + 1)}$

Step 4.3.4

Cancel the common factor of $- 4$ and $2$ .

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Step 4.3.4.1

Factor $2$ out of $- 4 x^{2}$ .

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

Step 4.3.4.2

Cancel the common factors.

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Step 4.3.4.2.1

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

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

Step 4.3.4.2.2

Cancel the common factor.

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

Step 4.3.4.2.3

Rewrite the expression.

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

Step 4.3.5

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

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Step 4.3.5.1

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

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

Step 4.3.5.2

Cancel the common factors.

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Step 4.3.5.2.1

Cancel the common factor.

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

Step 4.3.5.2.2

Rewrite the expression.

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

Step 4.3.6

Move the negative in front of the fraction.

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

Step 4.4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Multiply $2$ by $- 1$ .

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

Step 5.4

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.4.1

Let $u = x^{2} + 1$ . Find $\frac{d u}{d x}$ .

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Step 5.4.1.1

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

Step 5.4.1.2

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

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 5.4.1.3

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

$2 x + \frac{d}{d x} [1]$

Step 5.4.1.4

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

$2 x + 0$

Step 5.4.1.5

Add $2 x$ and $0$ .

$2 x$

Step 5.4.2

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

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

Step 5.5

Simplify.

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Step 5.5.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

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

Step 5.5.2

Move $2$ to the left of $u$ .

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

Step 5.6

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

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

Step 5.7

Simplify.

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Step 5.7.1

Combine $\frac{1}{2}$ and $- 2$ .

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

Step 5.7.2

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

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Step 5.7.2.1

Factor $2$ out of $- 2$ .

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

Step 5.7.2.2

Cancel the common factors.

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Step 5.7.2.2.1

Factor $2$ out of $2$ .

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

Step 5.7.2.2.2

Cancel the common factor.

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

Step 5.7.2.2.3

Rewrite the expression.

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

Step 5.7.2.2.4

Divide $- 1$ by $1$ .

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

Step 5.8

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

$μ (x, y) = e^{- (\ln (| u |) + C)}$

Step 5.9

Simplify.

$μ (x, y) = e^{- \ln (| u |)} + C$

Step 5.10

Replace all occurrences of $u$ with $x^{2} + 1$ .

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

Step 5.11

Simplify each term.

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Step 5.11.1

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

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

Step 5.11.2

Exponentiation and log are inverse functions.

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

Step 5.11.3

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

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

Step 6

Multiply both sides of $(2 x^{2} y + 2 y + 5) d x + (2 x^{3} + 2 x) d y = 0$ by the integration factor $\frac{1}{x^{2} + 1}$ .

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Step 6.1

Multiply $2 x^{2} y + 2 y + 5$ by $\frac{1}{x^{2} + 1}$ .

$(2 x^{2} y + 2 y + 5) \frac{1}{x^{2} + 1} d x + (2 x^{3} + 2 x) d y = 0$

Step 6.2

Multiply $2 x^{2} y + 2 y + 5$ by $\frac{1}{x^{2} + 1}$ .

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + (2 x^{3} + 2 x) d y = 0$

Step 6.3

Multiply $2 x^{3} + 2 x$ by $\frac{1}{x^{2} + 1}$ .

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + (2 x^{3} + 2 x) \frac{1}{x^{2} + 1} d y = 0$

Step 6.4

Multiply $2 x^{3} + 2 x$ by $\frac{1}{x^{2} + 1}$ .

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + \frac{2 x^{3} + 2 x}{x^{2} + 1} d y = 0$

Step 6.5

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

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Step 6.5.1

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

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + \frac{2 x (x^{2}) + 2 x}{x^{2} + 1} d y = 0$

Step 6.5.2

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

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + \frac{2 x (x^{2}) + 2 x (1)}{x^{2} + 1} d y = 0$

Step 6.5.3

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

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + \frac{2 x (x^{2} + 1)}{x^{2} + 1} d y = 0$

Step 6.6

Cancel the common factor of $x^{2} + 1$ .

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Step 6.6.1

Cancel the common factor.

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + \frac{2 x (x^{2} + 1)}{x^{2} + 1} d y = 0$

Step 6.6.2

Divide $2 x$ by $1$ .

$\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} d x + 2 x d y = 0$

Step 7

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

$f (x, y) = \int 2 x d y$

Step 8

Integrate $N (x, y) = 2 x$ to find $f (x, y)$ .

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Step 8.1

Apply the constant rule.

$f (x, y) = 2 x y + C$

Step 9

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

$f (x, y) = 2 x y + g (x)$

Step 10

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

$\frac{\partial f}{\partial x} = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [2 x y + g (x)] = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11.2

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

$\frac{d}{d x} [2 x y] + \frac{d}{d x} [g (x)] = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11.3

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

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Step 11.3.1

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

$2 y \frac{d}{d x} [x] + \frac{d}{d x} [g (x)] = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11.3.2

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

$2 y \cdot 1 + \frac{d}{d x} [g (x)] = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11.3.3

Multiply $2$ by $1$ .

$2 y + \frac{d}{d x} [g (x)] = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11.4

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

$2 y + \frac{d g}{d x} = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 11.5

Reorder terms.

$\frac{d g}{d x} + 2 y = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$

Step 12

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

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Step 12.1

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

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Step 12.1.1

Subtract $2 y$ from both sides of the equation.

$\frac{d g}{d x} = \frac{2 x^{2} y + 2 y + 5}{x^{2} + 1} - 2 y$

Step 12.1.2

Simplify each term.

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Step 12.1.2.1

Split the fraction $\frac{2 x^{2} y + 2 y + 5}{x^{2} + 1}$ into two fractions.

$\frac{d g}{d x} = \frac{2 x^{2} y + 2 y}{x^{2} + 1} + \frac{5}{x^{2} + 1} - 2 y$

Step 12.1.2.2

Simplify each term.

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Step 12.1.2.2.1

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

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Step 12.1.2.2.1.1

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

$\frac{d g}{d x} = \frac{2 y (x^{2}) + 2 y}{x^{2} + 1} + \frac{5}{x^{2} + 1} - 2 y$

Step 12.1.2.2.1.2

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

$\frac{d g}{d x} = \frac{2 y (x^{2}) + 2 y (1)}{x^{2} + 1} + \frac{5}{x^{2} + 1} - 2 y$

Step 12.1.2.2.1.3

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

$\frac{d g}{d x} = \frac{2 y (x^{2} + 1)}{x^{2} + 1} + \frac{5}{x^{2} + 1} - 2 y$

Step 12.1.2.2.2

Cancel the common factor of $x^{2} + 1$ .

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Step 12.1.2.2.2.1

Cancel the common factor.

$\frac{d g}{d x} = \frac{2 y (x^{2} + 1)}{x^{2} + 1} + \frac{5}{x^{2} + 1} - 2 y$

Step 12.1.2.2.2.2

Divide $2 y$ by $1$ .

$\frac{d g}{d x} = 2 y + \frac{5}{x^{2} + 1} - 2 y$

Step 12.1.3

Combine the opposite terms in $2 y + \frac{5}{x^{2} + 1} - 2 y$ .

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Step 12.1.3.1

Subtract $2 y$ from $2 y$ .

$\frac{d g}{d x} = 0 + \frac{5}{x^{2} + 1}$

Step 12.1.3.2

Add $0$ and $\frac{5}{x^{2} + 1}$ .

$\frac{d g}{d x} = \frac{5}{x^{2} + 1}$

Step 13

Find the antiderivative of $\frac{5}{x^{2} + 1}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = \frac{5}{x^{2} + 1}$ .

$\int \frac{d g}{d x} d x = \int \frac{5}{x^{2} + 1} d x$

Step 13.2

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

$g (x) = \int \frac{5}{x^{2} + 1} d x$

Step 13.3

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$g (x) = 5 \int \frac{1}{x^{2} + 1} d x$

Step 13.4

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

$g (x) = 5 \int \frac{1}{1 + x^{2}} d x$

Step 13.5

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

$g (x) = 5 \int \frac{1}{1^{2} + x^{2}} d x$

Step 13.6

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x) + C$ .

$g (x) = 5 (\arctan (x) + C)$

Step 13.7

Simplify.

$g (x) = 5 \arctan (x) + C$

Step 14

Substitute for $g (x)$ in $f (x, y) = 2 x y + g (x)$ .

$f (x, y) = 2 x y + 5 \arctan (x) + C$