Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Since $1$ is constant with respect to $y$ , the derivative of $1$ 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 $N (x, y) = - (1 + x^{2})$ .

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Step 2.1

Differentiate $N$ with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

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

Step 2.6

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

$\frac{\partial N}{\partial x} = - (2 x)$

Step 2.7

Multiply $2$ by $- 1$ .

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

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 $- 2 x$ for $\frac{\partial N}{\partial x}$ .

$x = - 2 x$

Step 3.2

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

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

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

Step 4.2

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

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

Step 4.3

Substitute $- (1 + x^{2})$ for $N$ .

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Step 4.3.1

Substitute $- (1 + x^{2})$ for $N$ .

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

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

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

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Step 4.3.2.1.1

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

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

Step 4.3.2.1.2

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

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

Step 4.3.2.1.3

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

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

Step 4.3.2.1.4

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

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

Step 4.3.2.2

Multiply $- 1$ by $- 2$ .

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

Step 4.3.2.3

Add $1$ and $2$ .

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

Step 4.3.3

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

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

Step 4.3.4

Move the negative in front of the fraction.

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

Step 4.4

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

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

Step 5

Evaluate the integral $e^{\int - \frac{3 x}{1 + x^{2}} d x}$ .

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Step 5.1

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

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

Step 5.2

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 5.3

Multiply $3$ by $- 1$ .

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

Step 5.4

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

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Step 5.4.1.1

Differentiate $1 + x^{2}$ .

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

Step 5.4.1.2

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

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

Step 5.4.1.3

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

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

Step 5.4.1.4

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

$0 + 2 x$

Step 5.4.1.5

Add $0$ and $2 x$ .

$2 x$

Step 5.4.2

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

$μ (x, y) = e^{- 3 \int \frac{1}{u} \cdot \frac{1}{2} d u}$

Step 5.5

Simplify.

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Step 5.5.1

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

$μ (x, y) = e^{- 3 \int \frac{1}{u \cdot 2} d u}$

Step 5.5.2

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

$μ (x, y) = e^{- 3 \int \frac{1}{2 u} d u}$

Step 5.6

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

$μ (x, y) = e^{- 3 (\frac{1}{2} \int \frac{1}{u} d u)}$

Step 5.7

Simplify.

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Step 5.7.1

Combine $\frac{1}{2}$ and $- 3$ .

$μ (x, y) = e^{\frac{- 3}{2} \int \frac{1}{u} d u}$

Step 5.7.2

Move the negative in front of the fraction.

$μ (x, y) = e^{- \frac{3}{2} \int \frac{1}{u} d u}$

Step 5.8

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

$μ (x, y) = e^{- \frac{3}{2} (\ln (| u |) + C)}$

Step 5.9

Simplify.

$μ (x, y) = e^{- \frac{3}{2} \ln (| u |)} + C$

Step 5.10

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

$μ (x, y) = e^{- \frac{3}{2} \ln (| 1 + x^{2} |)} + C$

Step 5.11

Simplify each term.

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Step 5.11.1

Multiply $- \frac{3}{2} \ln (| 1 + x^{2} |)$ .

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Step 5.11.1.1

Reorder $\ln (| 1 + x^{2} |)$ and $\frac{3}{2}$ .

$μ (x, y) = e^{- (\frac{3}{2} \ln (| 1 + x^{2} |))} + C$

Step 5.11.1.2

Simplify $\frac{3}{2} \ln (| 1 + x^{2} |)$ by moving $\frac{3}{2}$ inside the logarithm.

$μ (x, y) = e^{- \ln ({| 1 + x^{2} |}^{\frac{3}{2}})} + C$

Step 5.11.2

Simplify $- \ln ({| 1 + x^{2} |}^{\frac{3}{2}})$ by moving $- 1$ inside the logarithm.

$μ (x, y) = e^{\ln ({({| 1 + x^{2} |}^{\frac{3}{2}})}^{- 1})} + C$

Step 5.11.3

Exponentiation and log are inverse functions.

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

Step 5.11.4

Multiply the exponents in ${({| 1 + x^{2} |}^{\frac{3}{2}})}^{- 1}$ .

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Step 5.11.4.1

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

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

Step 5.11.4.2

Multiply $\frac{3}{2} \cdot - 1$ .

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Step 5.11.4.2.1

Combine $\frac{3}{2}$ and $- 1$ .

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

Step 5.11.4.2.2

Multiply $3$ by $- 1$ .

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

Step 5.11.4.3

Move the negative in front of the fraction.

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

Step 5.11.5

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

Apply the distributive property.

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

Step 6.5

Multiply $- 1$ by $1$ .

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

Step 6.6

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

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

Step 6.7

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

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

Step 6.8

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

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

Step 6.9

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

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

Step 6.10

Move the negative in front of the fraction.

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

Step 7

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

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

Step 8

Integrate $N (x, y) = - \frac{1 + x^{2}}{{(1 + x^{2})}^{\frac{3}{2}}}$ to find $f (x, y)$ .

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Step 8.1

Apply the constant rule.

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

Step 8.2

Simplify.

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Step 8.2.1

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

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

Step 8.2.2

Move ${(1 + x^{2})}^{1}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

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

Step 8.2.3

Multiply ${(1 + x^{2})}^{\frac{3}{2}}$ by ${(1 + x^{2})}^{- 1}$ by adding the exponents.

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Step 8.2.3.1

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

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

Step 8.2.3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 8.2.3.3

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

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

Step 8.2.3.4

Combine the numerators over the common denominator.

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

Step 8.2.3.5

Simplify the numerator.

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Step 8.2.3.5.1

Multiply $- 1$ by $2$ .

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

Step 8.2.3.5.2

Subtract $2$ from $3$ .

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

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

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Step 11.3.1

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

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

Step 11.3.2

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

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

Step 11.3.3

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

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Step 11.3.3.1

To apply the Chain Rule, set $u_{1}$ as ${(1 + x^{2})}^{\frac{1}{2}}$ .

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

Step 11.3.3.2

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

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

Step 11.3.3.3

Replace all occurrences of $u_{1}$ with ${(1 + x^{2})}^{\frac{1}{2}}$ .

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

Step 11.3.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 1 + x^{2}$ .

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Step 11.3.4.1

To apply the Chain Rule, set $u_{2}$ as $1 + x^{2}$ .

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

Step 11.3.4.2

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

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

Step 11.3.4.3

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

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

Step 11.3.5

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

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

Step 11.3.6

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

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

Step 11.3.7

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

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

Step 11.3.8

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

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Step 11.3.8.1

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

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

Step 11.3.8.2

Cancel the common factor of $2$ .

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Step 11.3.8.2.1

Factor $2$ out of $- 2$ .

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

Step 11.3.8.2.2

Cancel the common factor.

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

Step 11.3.8.2.3

Rewrite the expression.

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

Step 11.3.9

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 11.3.10

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

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

Step 11.3.11

Combine the numerators over the common denominator.

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

Step 11.3.12

Simplify the numerator.

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Step 11.3.12.1

Multiply $- 1$ by $2$ .

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

Step 11.3.12.2

Subtract $2$ from $1$ .

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

Step 11.3.13

Move the negative in front of the fraction.

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

Step 11.3.14

Add $0$ and $2 x$ .

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

Step 11.3.15

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

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

Step 11.3.16

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

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

Step 11.3.17

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

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

Step 11.3.18

Move ${(1 + x^{2})}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 11.3.19

Cancel the common factor.

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

Step 11.3.20

Rewrite the expression.

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

Step 11.3.21

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

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

Step 11.3.22

Move ${(1 + x^{2})}^{- 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 11.3.23

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

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Step 11.3.23.1

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

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Step 11.3.23.1.1

Raise $1 + x^{2}$ to the power of $1$ .

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

Step 11.3.23.1.2

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

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

Step 11.3.23.2

Write $1$ as a fraction with a common denominator.

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

Step 11.3.23.3

Combine the numerators over the common denominator.

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

Step 11.3.23.4

Add $1$ and $2$ .

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

Step 11.3.24

Multiply $- 1$ by $- 1$ .

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

Step 11.3.25

Multiply $y$ by $1$ .

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

Step 11.3.26

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

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

Step 11.4

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

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

Step 11.5

Reorder terms.

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

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

Move all terms containing variables to the left side of the equation.

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Step 12.1.1.1

Subtract $\frac{1 + x y}{{(1 + x^{2})}^{\frac{3}{2}}}$ from both sides of the equation.

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

Step 12.1.1.2

Combine the numerators over the common denominator.

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

Step 12.1.1.3

Simplify each term.

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Step 12.1.1.3.1

Apply the distributive property.

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

Step 12.1.1.3.2

Multiply $- 1$ by $1$ .

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

Step 12.1.1.4

Combine the opposite terms in $y x - 1 - x y$ .

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Step 12.1.1.4.1

Reorder the factors in the terms $y x$ and $- x y$ .

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

Step 12.1.1.4.2

Subtract $x y$ from $x y$ .

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

Step 12.1.1.4.3

Subtract $1$ from $0$ .

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

Step 12.1.1.5

Move the negative in front of the fraction.

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

Step 12.1.2

Add $\frac{1}{{(x^{2} + 1)}^{\frac{3}{2}}}$ to both sides of the equation.

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

Step 13

Find the antiderivative of $\frac{1}{{(x^{2} + 1)}^{\frac{3}{2}}}$ to find $g (x)$ .

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Step 13.1

Integrate both sides of $\frac{d g}{d x} = \frac{1}{{(x^{2} + 1)}^{\frac{3}{2}}}$ .

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

Step 13.2

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

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

Step 13.3

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

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

Step 13.4

Let $x = \tan (t)$ , where $- \frac{π}{2} \leq t \leq \frac{π}{2}$ . Then $d x = \sec^{2} (t) d t$ . Note that since $- \frac{π}{2} \leq t \leq \frac{π}{2}$ , $\sec^{2} (t)$ is positive.

$g (x) = \int \frac{1}{\sqrt{{(\tan^{2} (t) + 1)}^{3}}} \sec^{2} (t) d t$

Step 13.5

Simplify $\sqrt{{(\tan^{2} (t) + 1)}^{3}}$ .

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Step 13.5.1

Apply pythagorean identity.

$g (x) = \int \frac{1}{\sqrt{{(\sec^{2} (t))}^{3}}} \sec^{2} (t) d t$

Step 13.5.2

Multiply the exponents in ${(\sec^{2} (t))}^{3}$ .

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Step 13.5.2.1

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

$g (x) = \int \frac{1}{\sqrt{{\sec (t)}^{2 \cdot 3}}} \sec^{2} (t) d t$

Step 13.5.2.2

Multiply $2$ by $3$ .

$g (x) = \int \frac{1}{\sqrt{\sec^{6} (t)}} \sec^{2} (t) d t$

Step 13.5.3

Rewrite $\sec^{6} (t)$ as ${(\sec^{3} (t))}^{2}$ .

$g (x) = \int \frac{1}{\sqrt{{(\sec^{3} (t))}^{2}}} \sec^{2} (t) d t$

Step 13.5.4

Pull terms out from under the radical, assuming positive real numbers.

$g (x) = \int \frac{1}{\sec^{3} (t)} \sec^{2} (t) d t$

Step 13.6

Cancel the common factor of $\sec^{2} (t)$ .

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Step 13.6.1

Factor $\sec^{2} (t)$ out of $\sec^{3} (t)$ .

$g (x) = \int \frac{1}{\sec^{2} (t) \sec (t)} \sec^{2} (t) d t$

Step 13.6.2

Cancel the common factor.

$g (x) = \int \frac{1}{\sec^{2} (t) \sec (t)} \sec^{2} (t) d t$

Step 13.6.3

Rewrite the expression.

$g (x) = \int \frac{1}{\sec (t)} d t$

Step 13.7

Simplify.

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Step 13.7.1

Rewrite $\sec (t)$ in terms of sines and cosines.

$g (x) = \int \frac{1}{\frac{1}{\cos (t)}} d t$

Step 13.7.2

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (t)}$ .

$g (x) = \int 1 \cos (t) d t$

Step 13.7.3

Multiply $\cos (t)$ by $1$ .

$g (x) = \int \cos (t) d t$

Step 13.8

The integral of $\cos (t)$ with respect to $t$ is $\sin (t)$ .

$g (x) = \sin (t) + C$

Step 13.9

Replace all occurrences of $t$ with $\arctan (x)$ .

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

Step 14

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

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

Step 15

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

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Step 15.1

Simplify each term.

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Step 15.1.1

Draw a triangle in the plane with vertices $(1, x)$ , $(1, 0)$ , and the origin. Then $\arctan (x)$ is the angle between the positive x-axis and the ray beginning at the origin and passing through $(1, x)$ . Therefore, $\sin (\arctan (x))$ is $\frac{x}{\sqrt{1 + x^{2}}}$ .

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

Step 15.1.2

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

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

Step 15.1.3

Combine and simplify the denominator.

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Step 15.1.3.1

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

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

Step 15.1.3.2

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

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

Step 15.1.3.3

Raise $\sqrt{1 + x^{2}}$ to the power of $1$ .

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

Step 15.1.3.4

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

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

Step 15.1.3.5

Add $1$ and $1$ .

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

Step 15.1.3.6

Rewrite ${\sqrt{1 + x^{2}}}^{2}$ as $1 + x^{2}$ .

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Step 15.1.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + x^{2}}$ as ${(1 + x^{2})}^{\frac{1}{2}}$ .

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

Step 15.1.3.6.2

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

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

Step 15.1.3.6.3

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

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

Step 15.1.3.6.4

Cancel the common factor of $2$ .

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Step 15.1.3.6.4.1

Cancel the common factor.

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

Step 15.1.3.6.4.2

Rewrite the expression.

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

Step 15.1.3.6.5

Simplify.

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

Step 15.2

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

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

Step 15.3

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

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

Step 15.4

Write each expression with a common denominator of $(1 + x^{2}) {(1 + x^{2})}^{\frac{1}{2}}$ , by multiplying each by an appropriate factor of $1$ .

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Step 15.4.1

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

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

Step 15.4.2

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

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Step 15.4.2.1

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

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Step 15.4.2.1.1

Raise $1 + x^{2}$ to the power of $1$ .

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

Step 15.4.2.1.2

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

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

Step 15.4.2.2

Write $1$ as a fraction with a common denominator.

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

Step 15.4.2.3

Combine the numerators over the common denominator.

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

Step 15.4.2.4

Add $1$ and $2$ .

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

Step 15.4.3

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

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

Step 15.4.4

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

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Step 15.4.4.1

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

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Step 15.4.4.1.1

Raise $1 + x^{2}$ to the power of $1$ .

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

Step 15.4.4.1.2

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

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

Step 15.4.4.2

Write $1$ as a fraction with a common denominator.

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

Step 15.4.4.3

Combine the numerators over the common denominator.

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

Step 15.4.4.4

Add $2$ and $1$ .

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

Step 15.5

Combine the numerators over the common denominator.

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

Step 15.6

Simplify the numerator.

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Step 15.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + x^{2}}$ as ${(1 + x^{2})}^{\frac{1}{2}}$ .

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

Step 15.6.2

Apply the distributive property.

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

Step 15.6.3

Multiply $- 1$ by $1$ .

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

Step 15.6.4

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

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Step 15.6.4.1

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

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

Step 15.6.4.2

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

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

Step 15.6.4.3

Combine the numerators over the common denominator.

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

Step 15.6.4.4

Add $1$ and $1$ .

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

Step 15.6.4.5

Divide $2$ by $2$ .

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

Step 15.6.5

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

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

Step 15.6.6

Apply the distributive property.

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

Step 15.6.7

Multiply $x$ by $1$ .

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

Step 15.6.8

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

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Step 15.6.8.1

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

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Step 15.6.8.1.1

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

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

Step 15.6.8.1.2

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

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

Step 15.6.8.2

Add $1$ and $2$ .

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

Step 15.6.9

Rewrite $- y - y x^{2} + x + x^{3}$ in a factored form.

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Step 15.6.9.1

Factor out the greatest common factor from each group.

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Step 15.6.9.1.1

Group the first two terms and the last two terms.

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

Step 15.6.9.1.2

Factor out the greatest common factor (GCF) from each group.

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

Step 15.6.9.2

Factor the polynomial by factoring out the greatest common factor, $- 1 - x^{2}$ .

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