Calculus Examples

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

Step 1

Rewrite the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Cancel the common factor.

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

Step 3.2.3

Rewrite the expression.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

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 4.2.1.1

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

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Step 4.2.1.1.1

Differentiate $1 + y^{2}$ .

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

Step 4.2.1.1.2

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 4.2.1.1.3

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 4.2.1.1.4

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 4.2.1.1.5

Add $0$ and $2 y$ .

$2 y$

Step 4.2.1.2

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

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

Step 4.2.2

Simplify.

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Step 4.2.2.1

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

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

Step 4.2.2.2

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

$\int \frac{1}{2 u} d u = \int x d x$

Step 4.2.3

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

$\frac{1}{2} \int \frac{1}{u} d u = \int x d x$

Step 4.2.4

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

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

Step 4.2.5

Simplify.

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

Step 4.2.6

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

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

Step 4.3

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

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

Step 4.4

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

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

Step 5

Solve for $y$ .

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Step 5.1

Multiply both sides of the equation by $2$ .

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

Step 5.2

Simplify both sides of the equation.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

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

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Step 5.2.1.1.1

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

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

Step 5.2.1.1.2

Cancel the common factor of $2$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

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Step 5.2.2.1.1

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

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

Step 5.2.2.1.2

Apply the distributive property.

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

Step 5.2.2.1.3

Cancel the common factor of $2$ .

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Step 5.2.2.1.3.1

Cancel the common factor.

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

Step 5.2.2.1.3.2

Rewrite the expression.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Solve for $y$ .

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Step 5.5.1

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

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

Step 5.5.2

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

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

Step 5.5.3

Subtract $1$ from both sides of the equation.

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

Step 5.5.4

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

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

Step 6

Group the constant terms together.

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Step 6.1

Simplify the constant of integration.

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

Step 6.2

Rewrite $e^{x^{2} + C}$ as $e^{x^{2}} e^{C}$ .

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

Step 6.3

Reorder $e^{x^{2}}$ and $e^{C}$ .

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

Step 6.4

Combine constants with the plus or minus.

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