Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Regroup factors.

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

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

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

Step 1.3.2

Cancel the common factor of $y$ .

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Step 1.3.2.1

Factor $y$ out of $x y$ .

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

Step 1.3.2.2

Cancel the common factor.

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

Step 1.3.2.3

Rewrite the expression.

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

Step 1.3.3

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

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Step 1.3.3.1

Factor $y^{2} + 1$ out of $\ln (x) (y^{2} + 1)$ .

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

Step 1.3.3.2

Cancel the common factor.

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

Step 1.3.3.3

Rewrite the expression.

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

Step 1.4

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

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 $d u_{1}$ .

$\int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} = \int \frac{\ln (x)}{x} d x$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$\int \frac{1}{u_{1} \cdot 2} d u_{1} = \int \frac{\ln (x)}{x} d x$

Step 2.2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u_{2} = \ln (x)$ . Then $d u_{2} = \frac{1}{x} d x$ , so $x d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 2.3.1.1

Let $u_{2} = \ln (x)$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.3.1.1.1

Differentiate $\ln (x)$ .

$\frac{d}{d x} [\ln (x)]$

Step 2.3.1.1.2

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

$\frac{1}{x}$

Step 2.3.1.2

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

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

Step 2.3.2

By the Power Rule, the integral of $u_{2}$ with respect to $u_{2}$ is $\frac{1}{2} {u_{2}}^{2}$ .

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

Step 2.3.3

Replace all occurrences of $u_{2}$ with $\ln (x)$ .

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $2$ .

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

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

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

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.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (\frac{1}{2} \ln^{2} (x) + C)$ .

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Apply the distributive property.

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

Step 3.2.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.2.1.3.1

Cancel the common factor.

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

Step 3.2.2.1.3.2

Rewrite the expression.

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Subtract $1$ from both sides of the equation.

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

Step 3.5.4

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

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

Rewrite $e^{\ln^{2} (x) + C}$ as $e^{\ln^{2} (x)} e^{C}$ .

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

Step 4.3

Reorder $e^{\ln^{2} (x)}$ and $e^{C}$ .

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

Step 4.4

Combine constants with the plus or minus.

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