Calculus Examples

$x \frac{d y}{d x} = \frac{y - 1}{y + 1} - y$

Step 1

Separate the variables.

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Step 1.1

Divide each term in $x \frac{d y}{d x} = \frac{y - 1}{y + 1} - y$ by $x$ and simplify.

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Step 1.1.1

Divide each term in $x \frac{d y}{d x} = \frac{y - 1}{y + 1} - y$ by $x$ .

$\frac{x \frac{d y}{d x}}{x} = \frac{\frac{y - 1}{y + 1}}{x} + \frac{- y}{x}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $x$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{x \frac{d y}{d x}}{x} = \frac{\frac{y - 1}{y + 1}}{x} + \frac{- y}{x}$

Step 1.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

$\frac{d y}{d x} = \frac{\frac{y - 1}{y + 1}}{x} + \frac{- y}{x}$

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{\frac{y - 1}{y + 1} - y}{x}$

Step 1.1.3.2

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

$\frac{d y}{d x} = \frac{\frac{y - 1}{y + 1} - y \cdot \frac{y + 1}{y + 1}}{x}$

Step 1.1.3.3

Simplify terms.

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Step 1.1.3.3.1

Combine $- y$ and $\frac{y + 1}{y + 1}$ .

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

Step 1.1.3.3.2

Combine the numerators over the common denominator.

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

Step 1.1.3.4

Simplify the numerator.

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Step 1.1.3.4.1

Apply the distributive property.

$\frac{d y}{d x} = \frac{\frac{y - 1 - y \cdot y - y \cdot 1}{y + 1}}{x}$

Step 1.1.3.4.2

Multiply $y$ by $y$ by adding the exponents.

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Step 1.1.3.4.2.1

Move $y$ .

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

Step 1.1.3.4.2.2

Multiply $y$ by $y$ .

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

Step 1.1.3.4.3

Multiply $- 1$ by $1$ .

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

Step 1.1.3.4.4

Subtract $y$ from $y$ .

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

Step 1.1.3.4.5

Add $- 1 - y^{2}$ and $0$ .

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

Step 1.1.3.5

Simplify with factoring out.

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Step 1.1.3.5.1

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

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

Step 1.1.3.5.2

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

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

Step 1.1.3.5.3

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

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

Step 1.1.3.5.4

Move the negative in front of the fraction.

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

Step 1.1.3.6

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.1.3.7

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

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.4.2

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

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

Step 1.4.3

Cancel the common factor of $y + 1$ .

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Step 1.4.3.1

Move the leading negative in $- \frac{y + 1}{1 + y^{2}}$ into the numerator.

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

Step 1.4.3.2

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

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

Step 1.4.3.3

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

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

Step 1.4.3.4

Cancel the common factor.

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

Step 1.4.3.5

Rewrite the expression.

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

Step 1.4.4

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

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Step 1.4.4.1

Cancel the common factor.

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

Step 1.4.4.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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

Step 2.2.2

Split the single integral into multiple integrals.

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

Step 2.2.3

Let $u = 1 + y^{2}$ . 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 2.2.3.1

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

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Step 2.2.3.1.1

Differentiate $1 + y^{2}$ .

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

Step 2.2.3.1.2

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

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

Step 2.2.3.1.3

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

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

Step 2.2.3.1.4

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

$0 + 2 y$

Step 2.2.3.1.5

Add $0$ and $2 y$ .

$2 y$

Step 2.2.3.2

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

$\int \frac{1}{u} \cdot \frac{1}{2} d u + \int \frac{1}{1 + y^{2}} d y = \int - \frac{1}{x} d x$

Step 2.2.4

Simplify.

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Step 2.2.4.1

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

$\int \frac{1}{u \cdot 2} d u + \int \frac{1}{1 + y^{2}} d y = \int - \frac{1}{x} d x$

Step 2.2.4.2

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

$\int \frac{1}{2 u} d u + \int \frac{1}{1 + y^{2}} d y = \int - \frac{1}{x} d x$

Step 2.2.5

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

$\frac{1}{2} \int \frac{1}{u} d u + \int \frac{1}{1 + y^{2}} d y = \int - \frac{1}{x} d x$

Step 2.2.6

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

$\frac{1}{2} (\ln (| u |) + C_{1}) + \int \frac{1}{1 + y^{2}} d y = \int - \frac{1}{x} d x$

Step 2.2.7

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

$\frac{1}{2} (\ln (| u |) + C_{1}) + \int \frac{1}{1^{2} + y^{2}} d y = \int - \frac{1}{x} d x$

Step 2.2.8

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

$\frac{1}{2} (\ln (| u |) + C_{1}) + \arctan (y) + C_{2} = \int - \frac{1}{x} d x$

Step 2.2.9

Simplify.

$\frac{1}{2} \ln (| u |) + \arctan (y) + C_{3} = \int - \frac{1}{x} d x$

Step 2.2.10

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

$\frac{1}{2} \ln (| 1 + y^{2} |) + \arctan (y) + C_{3} = \int - \frac{1}{x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

$\frac{1}{2} \ln (| 1 + y^{2} |) + \arctan (y) + C_{3} = - \int \frac{1}{x} d x$

Step 2.3.2

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

$\frac{1}{2} \ln (| 1 + y^{2} |) + \arctan (y) + C_{3} = - (\ln (| x |) + C_{4})$

Step 2.3.3

Simplify.

$\frac{1}{2} \ln (| 1 + y^{2} |) + \arctan (y) + C_{3} = - \ln (| x |) + C_{4}$

Step 2.4

Group the constant of integration on the right side as $C$ .

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