Calculus Examples

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

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Subtract $(1 + x y) d y$ from both sides of the equation.

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

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = y^{2} + 1$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Add $2 y$ and $0$ .

$\frac{\partial M}{\partial y} = 2 y$

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

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

$\frac{\partial N}{\partial x} = - (y \frac{d}{d x} [x])$

Step 3.7

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

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

Step 3.8

Multiply $- 1$ by $1$ .

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

Step 4

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

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Step 4.1

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

$2 y = - y$

Step 4.2

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

$2 y = - y$ is not an identity.

Step 5

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 5.1

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

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

Step 5.2

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

$\frac{- y - (2 y)}{M}$

Step 5.3

Substitute $y^{2} + 1$ for $M$ .

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Step 5.3.1

Substitute $y^{2} + 1$ for $M$ .

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

Step 5.3.2

Simplify the numerator.

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Step 5.3.2.1

Factor $y$ out of $- y - 1 \cdot 2 y$ .

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Step 5.3.2.1.1

Factor $y$ out of $- y$ .

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

Step 5.3.2.1.2

Factor $y$ out of $- 1 \cdot 2 y$ .

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

Step 5.3.2.1.3

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

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

Step 5.3.2.2

Multiply $- 1$ by $2$ .

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

Step 5.3.2.3

Subtract $2$ from $- 1$ .

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

Step 5.3.3

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

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

Step 5.3.4

Substitute $y^{2} + 1$ for $M$ .

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

Step 5.4

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Multiply $3$ by $- 1$ .

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

Step 6.4

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

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Step 6.4.1

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

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Step 6.4.1.1

Differentiate $y^{2} + 1$ .

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

Step 6.4.1.2

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

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

Step 6.4.1.3

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

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

Step 6.4.1.4

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

$2 y + 0$

Step 6.4.1.5

Add $2 y$ and $0$ .

$2 y$

Step 6.4.2

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

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

Step 6.5

Simplify.

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Step 6.5.1

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

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

Step 6.5.2

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

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

Step 6.6

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 6.7

Simplify.

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Step 6.7.1

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

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

Step 6.7.2

Move the negative in front of the fraction.

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

Step 6.8

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

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

Step 6.9

Simplify.

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

Step 6.10

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

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

Step 6.11

Simplify each term.

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Step 6.11.1

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

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Step 6.11.1.1

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

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

Step 6.11.1.2

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

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

Step 6.11.2

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

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

Step 6.11.3

Exponentiation and log are inverse functions.

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

Step 6.11.4

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

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Step 6.11.4.1

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

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

Step 6.11.4.2

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

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Step 6.11.4.2.1

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

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

Step 6.11.4.2.2

Multiply $3$ by $- 1$ .

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

Step 6.11.4.3

Move the negative in front of the fraction.

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

Step 6.11.5

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

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

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

Subtract the exponent from the denominator from the exponent of the numerator for the same base

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

Step 7.4

Simplify each term.

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Step 7.4.1

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

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Step 7.4.1.1

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

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

Step 7.4.1.2

Multiply $3$ by $- 1$ .

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

Step 7.4.2

Move the negative in front of the fraction.

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

Step 7.5

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

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

Step 7.6

Combine the numerators over the common denominator.

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

Step 7.7

Subtract $3$ from $2$ .

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

Step 7.8

Move the negative in front of the fraction.

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

Step 7.9

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

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

Step 7.10

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

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

Step 7.11

Apply the distributive property.

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

Step 7.12

Multiply $- 1$ by $1$ .

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

Step 7.13

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

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

Step 7.14

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

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

Step 7.15

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

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

Step 7.16

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

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

Step 7.17

Move the negative in front of the fraction.

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

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 9.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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Step 12.3.1

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

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

Step 12.3.2

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

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

Step 12.3.3

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

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Step 12.3.3.1

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

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

Step 12.3.3.2

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

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

Step 12.3.3.3

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

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

Step 12.3.4

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

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Step 12.3.4.1

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

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

Step 12.3.4.2

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

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

Step 12.3.4.3

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

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

Step 12.3.5

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

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

Step 12.3.6

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

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

Step 12.3.7

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

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

Step 12.3.8

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

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Step 12.3.8.1

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

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

Step 12.3.8.2

Cancel the common factor of $2$ .

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Step 12.3.8.2.1

Factor $2$ out of $- 2$ .

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

Step 12.3.8.2.2

Cancel the common factor.

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

Step 12.3.8.2.3

Rewrite the expression.

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

Step 12.3.9

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

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

Step 12.3.10

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

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

Step 12.3.11

Combine the numerators over the common denominator.

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

Step 12.3.12

Simplify the numerator.

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Step 12.3.12.1

Multiply $- 1$ by $2$ .

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

Step 12.3.12.2

Subtract $2$ from $1$ .

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

Step 12.3.13

Move the negative in front of the fraction.

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

Step 12.3.14

Add $2 y$ and $0$ .

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

Step 12.3.15

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

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

Step 12.3.16

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

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

Step 12.3.17

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

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

Step 12.3.18

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

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

Step 12.3.19

Cancel the common factor.

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

Step 12.3.20

Rewrite the expression.

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

Step 12.3.21

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

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

Step 12.3.22

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

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

Step 12.3.23

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

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Step 12.3.23.1

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

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Step 12.3.23.1.1

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

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

Step 12.3.23.1.2

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

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

Step 12.3.23.2

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

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

Step 12.3.23.3

Combine the numerators over the common denominator.

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

Step 12.3.23.4

Add $1$ and $2$ .

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

Step 12.3.24

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

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

Step 12.4

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

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

Step 12.5

Reorder terms.

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

Step 13

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

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Step 13.1

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

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Step 13.1.1

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

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Step 13.1.1.1

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

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

Step 13.1.1.2

Combine the numerators over the common denominator.

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

Step 13.1.1.3

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

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Step 13.1.1.3.1

Add $- x y$ and $x y$ .

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

Step 13.1.1.3.2

Add $0$ and $1$ .

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

Step 13.1.2

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

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

Step 14

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

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Step 14.1

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

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

Step 14.2

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

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

Step 14.3

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

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

Step 14.4

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

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

Step 14.5

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

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

Step 14.6

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

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Step 14.6.1

Apply pythagorean identity.

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

Step 14.6.2

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

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Step 14.6.2.1

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

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

Step 14.6.2.2

Multiply $2$ by $3$ .

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

Step 14.6.3

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

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

Step 14.6.4

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

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

Step 14.7

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

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Step 14.7.1

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

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

Step 14.7.2

Cancel the common factor.

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

Step 14.7.3

Rewrite the expression.

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

Step 14.8

Simplify.

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Step 14.8.1

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

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

Step 14.8.2

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

$g (y) = - \int 1 \cos (t) d t$

Step 14.8.3

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

$g (y) = - \int \cos (t) d t$

Step 14.9

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

$g (y) = - (\sin (t) + C)$

Step 14.10

Simplify.

$g (y) = - \sin (t) + C$

Step 14.11

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

$g (y) = - \sin (\arctan (y)) + C$

Step 15

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

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

Step 16

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

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Step 16.1

Simplify each term.

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Step 16.1.1

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

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

Step 16.1.2

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

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

Step 16.1.3

Combine and simplify the denominator.

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Step 16.1.3.1

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

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

Step 16.1.3.2

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

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

Step 16.1.3.3

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

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

Step 16.1.3.4

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

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

Step 16.1.3.5

Add $1$ and $1$ .

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

Step 16.1.3.6

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

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Step 16.1.3.6.1

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

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

Step 16.1.3.6.2

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

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

Step 16.1.3.6.3

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

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

Step 16.1.3.6.4

Cancel the common factor of $2$ .

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Step 16.1.3.6.4.1

Cancel the common factor.

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

Step 16.1.3.6.4.2

Rewrite the expression.

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

Step 16.1.3.6.5

Simplify.

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

Step 16.2

Reorder terms.

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

Step 16.3

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

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

Step 16.4

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

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

Step 16.5

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

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Step 16.5.1

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

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

Step 16.5.2

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

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Step 16.5.2.1

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

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Step 16.5.2.1.1

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

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

Step 16.5.2.1.2

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

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

Step 16.5.2.2

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

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

Step 16.5.2.3

Combine the numerators over the common denominator.

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

Step 16.5.2.4

Add $1$ and $2$ .

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

Step 16.5.3

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

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

Step 16.5.4

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

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Step 16.5.4.1

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

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Step 16.5.4.1.1

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

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

Step 16.5.4.1.2

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

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

Step 16.5.4.2

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

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

Step 16.5.4.3

Combine the numerators over the common denominator.

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

Step 16.5.4.4

Add $2$ and $1$ .

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

Step 16.6

Combine the numerators over the common denominator.

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

Step 16.7

Simplify each term.

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Step 16.7.1

Simplify the numerator.

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Step 16.7.1.1

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

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

Step 16.7.1.2

Apply the distributive property.

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

Step 16.7.1.3

Multiply $x$ by $1$ .

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

Step 16.7.1.4

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

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Step 16.7.1.4.1

Reorder terms.

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

Step 16.7.1.4.2

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

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Step 16.7.1.4.2.1

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

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

Step 16.7.1.4.2.2

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

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

Step 16.7.1.4.2.3

Combine the numerators over the common denominator.

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

Step 16.7.1.4.2.4

Add $1$ and $1$ .

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

Step 16.7.1.4.2.5

Divide $2$ by $2$ .

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

Step 16.7.1.4.3

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

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

Step 16.7.1.5

Apply the distributive property.

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

Step 16.7.1.6

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

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Step 16.7.1.6.1

Move $y^{2}$ .

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

Step 16.7.1.6.2

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

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Step 16.7.1.6.2.1

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

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

Step 16.7.1.6.2.2

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

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

Step 16.7.1.6.3

Add $2$ and $1$ .

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

Step 16.7.1.7

Multiply $- 1$ by $1$ .

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

Step 16.7.1.8

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

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Step 16.7.1.8.1

Factor out the greatest common factor from each group.

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Step 16.7.1.8.1.1

Group the first two terms and the last two terms.

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

Step 16.7.1.8.1.2

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

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

Step 16.7.1.8.2

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

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

Step 16.7.2

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

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

Step 16.7.3

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

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Step 16.7.3.1

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

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

Step 16.7.3.2

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

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

Step 16.7.3.3

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

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

Step 16.7.3.4

Combine the numerators over the common denominator.

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

Step 16.7.3.5

Simplify the numerator.

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Step 16.7.3.5.1

Multiply $- 1$ by $2$ .

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

Step 16.7.3.5.2

Subtract $2$ from $3$ .

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