Calculus Examples

$(\arctan (x) + x y) d x + (e^{y} + \frac{x^{2}}{2}) d y = 0$

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [\arctan (x) + x y]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{\partial M}{\partial y} = \frac{d}{d y} [\arctan (x)] + \frac{d}{d y} [x y]$

Step 1.2.2

Since $\arctan (x)$ is constant with respect to $y$ , the derivative of $\arctan (x)$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [x y]$

Step 1.3

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

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Step 1.3.1

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

$\frac{\partial M}{\partial y} = 0 + x \frac{d}{d y} [y]$

Step 1.3.2

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

$\frac{\partial M}{\partial y} = 0 + x \cdot 1$

Step 1.3.3

Multiply $x$ by $1$ .

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

Step 1.4

Add $0$ and $x$ .

$\frac{\partial M}{\partial y} = x$

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

Since $e^{y}$ is constant with respect to $x$ , the derivative of $e^{y}$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [\frac{x^{2}}{2}]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 0 + \frac{1}{2} \frac{d}{d x} [x^{2}]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 0 + \frac{1}{2} (2 x)$

Step 2.3.3

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

$\frac{\partial N}{\partial x} = 0 + \frac{2}{2} x$

Step 2.3.4

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

$\frac{\partial N}{\partial x} = 0 + \frac{2 x}{2}$

Step 2.3.5

Cancel the common factor of $2$ .

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Step 2.3.5.1

Cancel the common factor.

$\frac{\partial N}{\partial x} = 0 + \frac{2 x}{2}$

Step 2.3.5.2

Divide $x$ by $1$ .

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

Step 2.4

Add $0$ and $x$ .

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

Step 3

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

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Step 3.1

Substitute $x$ for $\frac{\partial M}{\partial y}$ and $x$ for $\frac{\partial N}{\partial x}$ .

$x = x$

Step 3.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$x = x$ is an identity.

Step 4

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

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

Step 5

Integrate $N (x, y) = e^{y} + \frac{x^{2}}{2}$ to find $f (x, y)$ .

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Step 5.1

Split the single integral into multiple integrals.

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

Step 5.2

The integral of $e^{y}$ with respect to $y$ is $e^{y}$ .

$f (x, y) = e^{y} + C + \int \frac{x^{2}}{2} d y$

Step 5.3

Apply the constant rule.

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

Step 5.4

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

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

Step 5.5

Simplify.

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

Step 6

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

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

Step 7

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

$\frac{\partial f}{\partial x} = \arctan (x) + x y$

Step 8

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

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Step 8.1

Differentiate $f$ with respect to $x$ .

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

Step 8.2

Differentiate.

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Step 8.2.1

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

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

Step 8.2.2

Since $e^{y}$ is constant with respect to $x$ , the derivative of $e^{y}$ with respect to $x$ is $0$ .

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

Step 8.3

Evaluate $\frac{d}{d x} [\frac{1}{2} x^{2} y]$ .

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Step 8.3.1

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

$0 + \frac{d}{d x} [\frac{x^{2}}{2} y] + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.2

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

$0 + \frac{d}{d x} [\frac{x^{2} y}{2}] + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.3

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

$0 + \frac{y}{2} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.4

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

$0 + \frac{y}{2} (2 x) + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.5

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

$0 + \frac{2 y}{2} x + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.6

Combine $\frac{2 y}{2}$ and $x$ .

$0 + \frac{2 y x}{2} + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.7

Cancel the common factor of $2$ .

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Step 8.3.7.1

Cancel the common factor.

$0 + \frac{2 y x}{2} + \frac{d}{d x} [g (x)] = \arctan (x) + x y$

Step 8.3.7.2

Divide $y x$ by $1$ .

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

Step 8.4

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

$0 + y x + \frac{d g}{d x} = \arctan (x) + x y$

Step 8.5

Simplify.

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Step 8.5.1

Add $0$ and $y x$ .

$y x + \frac{d g}{d x} = \arctan (x) + x y$

Step 8.5.2

Reorder terms.

$\frac{d g}{d x} + x y = \arctan (x) + x y$

Step 9

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

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Step 9.1

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

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Step 9.1.1

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

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

Step 9.1.2

Combine the opposite terms in $\arctan (x) + x y - x y$ .

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Step 9.1.2.1

Subtract $x y$ from $x y$ .

$\frac{d g}{d x} = \arctan (x) + 0$

Step 9.1.2.2

Add $\arctan (x)$ and $0$ .

$\frac{d g}{d x} = \arctan (x)$

Step 10

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

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Step 10.1

Integrate both sides of $\frac{d g}{d x} = \arctan (x)$ .

$\int \frac{d g}{d x} d x = \int \arctan (x) d x$

Step 10.2

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

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

Step 10.3

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \arctan (x)$ and $d v = 1$ .

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

Step 10.4

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

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

Step 10.5

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 10.5.1

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

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Step 10.5.1.1

Differentiate $x^{2} + 1$ .

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

Step 10.5.1.2

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 10.5.1.3

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 10.5.1.4

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

$2 x + 0$

Step 10.5.1.5

Add $2 x$ and $0$ .

$2 x$

Step 10.5.2

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

$g (x) = \arctan (x) x - \int \frac{1}{u} \cdot \frac{1}{2} d u$

Step 10.6

Simplify.

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Step 10.6.1

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

$g (x) = \arctan (x) x - \int \frac{1}{u \cdot 2} d u$

Step 10.6.2

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

$g (x) = \arctan (x) x - \int \frac{1}{2 \cdot u} d u$

Step 10.6.3

Multiply $2$ by $u$ .

$g (x) = \arctan (x) x - \int \frac{1}{2 u} d u$

Step 10.7

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

$g (x) = \arctan (x) x - (\frac{1}{2} \int \frac{1}{u} d u)$

Step 10.8

Remove parentheses.

$g (x) = \arctan (x) x - \frac{1}{2} \int \frac{1}{u} d u$

Step 10.9

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

$g (x) = \arctan (x) x - \frac{1}{2} (\ln (| u |) + C)$

Step 10.10

Simplify.

$g (x) = \arctan (x) x - \frac{1}{2} \ln (| u |) + C$

Step 10.11

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Simplify each term.

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Step 12.1.1

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

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

Step 12.1.2

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

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

Step 12.1.3

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

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Step 12.1.3.1

Reorder $\ln (| x^{2} + 1 |)$ and $\frac{1}{2}$ .

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

Step 12.1.3.2

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

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

Step 12.2

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

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

Step 12.3

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

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

Step 12.4

Combine the numerators over the common denominator.

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

Step 12.5

Simplify the numerator.

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Step 12.5.1

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

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Step 12.5.1.1

Multiply $2$ by $- 1$ .

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

Step 12.5.1.2

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

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

Step 12.5.2

Multiply the exponents in ${({| x^{2} + 1 |}^{\frac{1}{2}})}^{2}$ .

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Step 12.5.2.1

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

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

Step 12.5.2.2

Cancel the common factor of $2$ .

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Step 12.5.2.2.1

Cancel the common factor.

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

Step 12.5.2.2.2

Rewrite the expression.

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

Step 12.5.3

Simplify.

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

Step 12.6

Reorder factors in $f (x, y) = e^{y} + \arctan (x) x + \frac{x^{2} y - \ln (| x^{2} + 1 |)}{2} + C$ .

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