Calculus Examples

$x (1 + x^{2}) d x + y (1 + y^{2}) d y = 0$

Step 1

Subtract $x (1 + x^{2}) d x$ from both sides of the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify.

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Step 2.2.1.1

Apply the distributive property.

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

Step 2.2.1.2

Reorder $y$ and $1$ .

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

Step 2.2.1.3

Multiply $y$ by $1$ .

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

Step 2.2.1.4

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

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

Step 2.2.1.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.2.1.6

Add $1$ and $2$ .

$\int y + y^{3} d y = \int - x (1 + x^{2}) d x$

Step 2.2.1.7

Reorder $y$ and $y^{3}$ .

$\int y^{3} + y d y = \int - x (1 + x^{2}) d x$

Step 2.2.2

Split the single integral into multiple integrals.

$\int y^{3} d y + \int y d y = \int - x (1 + x^{2}) d x$

Step 2.2.3

By the Power Rule, the integral of $y^{3}$ with respect to $y$ is $\frac{1}{4} y^{4}$ .

$\frac{1}{4} y^{4} + C_{1} + \int y d y = \int - x (1 + x^{2}) d x$

Step 2.2.4

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

$\frac{1}{4} y^{4} + C_{1} + \frac{1}{2} y^{2} + C_{2} = \int - x (1 + x^{2}) d x$

Step 2.2.5

Simplify.

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = \int - x (1 + x^{2}) d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1.1

Differentiate $1 + x^{2}$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$0 + 2 x$

Step 2.3.1.1.5

Add $0$ and $2 x$ .

$2 x$

Step 2.3.1.2

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

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = \int - u \frac{1}{2} d u$

Step 2.3.2

Combine $\frac{1}{2}$ and $u$ .

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = \int - \frac{u}{2} d u$

Step 2.3.3

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

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = - \int \frac{u}{2} d u$

Step 2.3.4

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

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = - (\frac{1}{2} \int u d u)$

Step 2.3.5

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

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = - \frac{1}{2} (\frac{1}{2} u^{2} + C_{4})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Rewrite $- \frac{1}{2} (\frac{1}{2} u^{2} + C_{4})$ as $- \frac{1}{2} \cdot \frac{1}{2} u^{2} + C_{4}$ .

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = - \frac{1}{2} \cdot \frac{1}{2} u^{2} + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

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

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = - \frac{1}{2 \cdot 2} u^{2} + C_{4}$

Step 2.3.6.2.2

Multiply $2$ by $2$ .

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} + C_{3} = - \frac{1}{4} u^{2} + C_{4}$

Step 2.3.7

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

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

Step 2.4

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

$\frac{1}{4} y^{4} + \frac{1}{2} y^{2} = - \frac{1}{4} {(1 + x^{2})}^{2} + K$