Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Cancel the common factor of $y$ .

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Step 1.3.1.1

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

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

Step 1.3.1.2

Cancel the common factor.

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

Step 1.3.1.3

Rewrite the expression.

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

Step 1.3.2

Simplify the denominator.

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Step 1.3.2.1

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

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

Step 1.3.2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = (x + 1) (x - 1)$ . 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 2.3.1.1

Let $u = (x + 1) (x - 1)$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $(x + 1) (x - 1)$ .

$\frac{d}{d x} [(x + 1) (x - 1)]$

Step 2.3.1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 1$ and $g (x) = x - 1$ .

$(x + 1) \frac{d}{d x} [x - 1] + (x - 1) \frac{d}{d x} [x + 1]$

Step 2.3.1.1.3

Differentiate.

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Step 2.3.1.1.3.1

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

$(x + 1) (\frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 2.3.1.1.3.2

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

$(x + 1) (1 + \frac{d}{d x} [- 1]) + (x - 1) \frac{d}{d x} [x + 1]$

Step 2.3.1.1.3.3

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

$(x + 1) (1 + 0) + (x - 1) \frac{d}{d x} [x + 1]$

Step 2.3.1.1.3.4

Simplify the expression.

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Step 2.3.1.1.3.4.1

Add $1$ and $0$ .

$(x + 1) \cdot 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 2.3.1.1.3.4.2

Multiply $x + 1$ by $1$ .

$x + 1 + (x - 1) \frac{d}{d x} [x + 1]$

Step 2.3.1.1.3.5

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

$x + 1 + (x - 1) (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 2.3.1.1.3.6

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

$x + 1 + (x - 1) (1 + \frac{d}{d x} [1])$

Step 2.3.1.1.3.7

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

$x + 1 + (x - 1) (1 + 0)$

Step 2.3.1.1.3.8

Simplify by adding terms.

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Step 2.3.1.1.3.8.1

Add $1$ and $0$ .

$x + 1 + (x - 1) \cdot 1$

Step 2.3.1.1.3.8.2

Multiply $x - 1$ by $1$ .

$x + 1 + x - 1$

Step 2.3.1.1.3.8.3

Add $x$ and $x$ .

$2 x + 1 - 1$

Step 2.3.1.1.3.8.4

Simplify by subtracting numbers.

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Step 2.3.1.1.3.8.4.1

Subtract $1$ from $1$ .

$2 x + 0$

Step 2.3.1.1.3.8.4.2

Add $2 x$ and $0$ .

$2 x$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

$\ln (| y |) + C_{1} = \frac{1}{2} (\ln (| u |) + C_{2})$

Step 2.3.5

Simplify.

$\ln (| y |) + C_{1} = \frac{1}{2} \ln (| u |) + C_{2}$

Step 2.3.6

Replace all occurrences of $u$ with $(x + 1) (x - 1)$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Simplify the right side.

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Step 3.1.1

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

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

Step 3.2

Move all the terms containing a logarithm to the left side of the equation.

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

Step 3.3

Simplify the numerator.

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Step 3.3.1

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 3.3.1.1

Apply the distributive property.

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

Step 3.3.1.2

Apply the distributive property.

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

Step 3.3.1.3

Apply the distributive property.

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

Step 3.3.2

Simplify and combine like terms.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $x$ by $x$ .

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

Step 3.3.2.1.2

Move $- 1$ to the left of $x$ .

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

Step 3.3.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 3.3.2.1.4

Multiply $x$ by $1$ .

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

Step 3.3.2.1.5

Multiply $- 1$ by $1$ .

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

Step 3.3.2.2

Add $- x$ and $x$ .

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

Step 3.3.2.3

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

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

Step 3.4

To write $\ln (| y |)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.5

Simplify terms.

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Step 3.5.1

Combine $\ln (| y |)$ and $\frac{2}{2}$ .

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

Step 3.5.2

Combine the numerators over the common denominator.

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

Step 3.6

Move $2$ to the left of $\ln (| y |)$ .

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

Step 3.7

Simplify the left side.

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Step 3.7.1

Simplify $\frac{2 \ln (| y |) - \ln (| x^{2} - 1 |)}{2}$ .

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Step 3.7.1.1

Simplify the numerator.

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Step 3.7.1.1.1

Simplify $2 \ln (| y |)$ by moving $2$ inside the logarithm.

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

Step 3.7.1.1.2

Remove the absolute value in ${| y |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.7.1.1.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 3.7.1.1.4

Simplify the denominator.

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Step 3.7.1.1.4.1

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

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

Step 3.7.1.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 3.7.1.1.4.3

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 3.7.1.1.4.3.1

Apply the distributive property.

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

Step 3.7.1.1.4.3.2

Apply the distributive property.

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

Step 3.7.1.1.4.3.3

Apply the distributive property.

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

Step 3.7.1.1.4.4

Simplify and combine like terms.

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Step 3.7.1.1.4.4.1

Simplify each term.

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Step 3.7.1.1.4.4.1.1

Multiply $x$ by $x$ .

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

Step 3.7.1.1.4.4.1.2

Move $- 1$ to the left of $x$ .

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

Step 3.7.1.1.4.4.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 3.7.1.1.4.4.1.4

Multiply $x$ by $1$ .

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

Step 3.7.1.1.4.4.1.5

Multiply $- 1$ by $1$ .

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

Step 3.7.1.1.4.4.2

Add $- x$ and $x$ .

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

Step 3.7.1.1.4.4.3

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

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

Step 3.7.1.1.4.5

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

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

Step 3.7.1.1.4.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 3.7.1.2

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

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

Step 3.7.1.3

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

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

Step 3.7.1.4

Apply the product rule to $\frac{y^{2}}{| (x + 1) (x - 1) |}$ .

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

Step 3.7.1.5

Simplify the numerator.

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Step 3.7.1.5.1

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

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Step 3.7.1.5.1.1

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

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

Step 3.7.1.5.1.2

Cancel the common factor of $2$ .

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Step 3.7.1.5.1.2.1

Cancel the common factor.

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

Step 3.7.1.5.1.2.2

Rewrite the expression.

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

Step 3.7.1.5.2

Simplify.

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

Step 3.7.1.6

Simplify the denominator.

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Step 3.7.1.6.1

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 3.7.1.6.1.1

Apply the distributive property.

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

Step 3.7.1.6.1.2

Apply the distributive property.

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

Step 3.7.1.6.1.3

Apply the distributive property.

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

Step 3.7.1.6.2

Simplify and combine like terms.

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Step 3.7.1.6.2.1

Simplify each term.

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Step 3.7.1.6.2.1.1

Multiply $x$ by $x$ .

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

Step 3.7.1.6.2.1.2

Move $- 1$ to the left of $x$ .

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

Step 3.7.1.6.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 3.7.1.6.2.1.4

Multiply $x$ by $1$ .

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

Step 3.7.1.6.2.1.5

Multiply $- 1$ by $1$ .

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

Step 3.7.1.6.2.2

Add $- x$ and $x$ .

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

Step 3.7.1.6.2.3

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

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

Step 3.8

To solve for $y$ , rewrite the equation using properties of logarithms.

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

Step 3.9

Rewrite $\ln (\frac{y}{{| x^{2} - 1 |}^{\frac{1}{2}}}) = C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{C} = \frac{y}{{| x^{2} - 1 |}^{\frac{1}{2}}}$

Step 3.10

Solve for $y$ .

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Step 3.10.1

Rewrite the equation as $\frac{y}{{| x^{2} - 1 |}^{\frac{1}{2}}} = e^{C}$ .

$\frac{y}{{| x^{2} - 1 |}^{\frac{1}{2}}} = e^{C}$

Step 3.10.2

Multiply both sides by ${| x^{2} - 1 |}^{\frac{1}{2}}$ .

$\frac{y}{{| x^{2} - 1 |}^{\frac{1}{2}}} {| x^{2} - 1 |}^{\frac{1}{2}} = e^{C} {| x^{2} - 1 |}^{\frac{1}{2}}$

Step 3.10.3

Simplify the left side.

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Step 3.10.3.1

Cancel the common factor of ${| x^{2} - 1 |}^{\frac{1}{2}}$ .

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Step 3.10.3.1.1

Cancel the common factor.

$\frac{y}{{| x^{2} - 1 |}^{\frac{1}{2}}} {| x^{2} - 1 |}^{\frac{1}{2}} = e^{C} {| x^{2} - 1 |}^{\frac{1}{2}}$

Step 3.10.3.1.2

Rewrite the expression.

$y = e^{C} {| x^{2} - 1 |}^{\frac{1}{2}}$

Step 4

Simplify the constant of integration.

$y = C {| x^{2} - 1 |}^{\frac{1}{2}}$