Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Factor $y$ out of $x y + y$ .

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Step 1.1.1

Factor $y$ out of $x y$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

Factor $y$ out of $y x + y \cdot 1$ .

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

Step 1.2

Regroup factors.

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Cancel the common factor of $y$ .

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Step 1.4.2.1

Cancel the common factor.

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

Step 1.4.2.2

Rewrite the expression.

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

Step 1.4.3

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

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Step 1.4.3.1

Cancel the common factor.

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

Step 1.4.3.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y^{2} + 1$ .

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

Step 2.2.1.1.2

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 2.2.1.1.3

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 2.2.1.1.4

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

$2 y + 0$

Step 2.2.1.1.5

Add $2 y$ and $0$ .

$2 y$

Step 2.2.1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int \frac{1}{x + 1} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int \frac{1}{x + 1} d x$

Step 2.2.2.2

Move $2$ to the left of $u_{1}$ .

$\int \frac{1}{2 u_{1}} d u_{1} = \int \frac{1}{x + 1} d x$

Step 2.2.3

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

$\frac{1}{2} \int \frac{1}{u_{1}} d u_{1} = \int \frac{1}{x + 1} d x$

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = x + 1$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $x + 1$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$1 + 0$

Step 2.3.1.1.5

Add $1$ and $0$ .

$1$

Step 2.3.1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

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

Step 2.3.2

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

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

Step 2.3.3

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

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

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Apply the distributive property.

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

Step 3.3

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

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

Step 3.4

Simplify the left side.

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Step 3.4.1

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

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

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

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

Step 3.4.1.1.2

Remove the absolute value in ${| x + 1 |}^{2}$ because exponentiations with even powers are always positive.

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

Step 3.4.1.2

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

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

Solve for $y$ .

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Step 3.7.1

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

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

Step 3.7.2

Multiply both sides by ${(x + 1)}^{2}$ .

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

Step 3.7.3

Simplify the left side.

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Step 3.7.3.1

Cancel the common factor of ${(x + 1)}^{2}$ .

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Step 3.7.3.1.1

Cancel the common factor.

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

Step 3.7.3.1.2

Rewrite the expression.

$| y^{2} + 1 | = e^{2 C} {(x + 1)}^{2}$

Step 3.7.4

Solve for $y$ .

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Step 3.7.4.1

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

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Step 3.7.4.1.1

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

$| y^{2} + 1 | = e^{2 C} ((x + 1) (x + 1))$

Step 3.7.4.1.2

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

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Step 3.7.4.1.2.1

Apply the distributive property.

$| y^{2} + 1 | = e^{2 C} (x (x + 1) + 1 (x + 1))$

Step 3.7.4.1.2.2

Apply the distributive property.

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

Step 3.7.4.1.2.3

Apply the distributive property.

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

Step 3.7.4.1.3

Simplify and combine like terms.

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Step 3.7.4.1.3.1

Simplify each term.

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Step 3.7.4.1.3.1.1

Multiply $x$ by $x$ .

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

Step 3.7.4.1.3.1.2

Multiply $x$ by $1$ .

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

Step 3.7.4.1.3.1.3

Multiply $x$ by $1$ .

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

Step 3.7.4.1.3.1.4

Multiply $1$ by $1$ .

$| y^{2} + 1 | = e^{2 C} (x^{2} + x + x + 1)$

Step 3.7.4.1.3.2

Add $x$ and $x$ .

$| y^{2} + 1 | = e^{2 C} (x^{2} + 2 x + 1)$

Step 3.7.4.1.4

Apply the distributive property.

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

Step 3.7.4.1.5

Simplify.

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Step 3.7.4.1.5.1

Rewrite using the commutative property of multiplication.

$| y^{2} + 1 | = e^{2 C} x^{2} + 2 e^{2 C} x + e^{2 C} \cdot 1$

Step 3.7.4.1.5.2

Multiply $e^{2 C}$ by $1$ .

$| y^{2} + 1 | = e^{2 C} x^{2} + 2 e^{2 C} x + e^{2 C}$

Step 3.7.4.1.6

Reorder factors in $e^{2 C} x^{2} + 2 e^{2 C} x + e^{2 C}$ .

$| y^{2} + 1 | = x^{2} e^{2 C} + 2 x e^{2 C} + e^{2 C}$

Step 3.7.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y^{2} + 1 = \pm (x^{2} e^{2 C} + 2 x e^{2 C} + e^{2 C})$

Step 3.7.4.3

Subtract $1$ from both sides of the equation.

$y^{2} = \pm (x^{2} e^{2 C} + 2 x e^{2 C} + e^{2 C}) - 1$

Step 3.7.4.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\pm (x^{2} e^{2 C} + 2 x e^{2 C} + e^{2 C}) - 1}$

Step 4

Simplify the constant of integration.

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