Calculus Examples

$x (1 + 3 y^{2}) d y - (1 + 2 x^{2}) d x = 0$

Step 1

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

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $x$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

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

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int 1 + 3 y^{2} d y = \int \frac{1 + 2 x^{2}}{x} d x$

Step 4.2

Integrate the left side.

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Step 4.2.1

Split the single integral into multiple integrals.

$\int d y + \int 3 y^{2} d y = \int \frac{1 + 2 x^{2}}{x} d x$

Step 4.2.2

Apply the constant rule.

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

Step 4.2.3

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

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

Step 4.2.4

By the Power Rule, the integral of $y^{2}$ with respect to $y$ is $\frac{1}{3} y^{3}$ .

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

Step 4.2.5

Simplify.

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Step 4.2.5.1

Simplify.

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

Step 4.2.5.2

Simplify.

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Step 4.2.5.2.1

Combine $3$ and $\frac{1}{3}$ .

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

Step 4.2.5.2.2

Cancel the common factor of $3$ .

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Step 4.2.5.2.2.1

Cancel the common factor.

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

Step 4.2.5.2.2.2

Rewrite the expression.

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

Step 4.2.5.2.3

Multiply $y^{3}$ by $1$ .

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

Step 4.2.6

Reorder terms.

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

Step 4.3

Integrate the right side.

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Step 4.3.1

Split the fraction into multiple fractions.

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

Step 4.3.2

Split the single integral into multiple integrals.

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

Step 4.3.3

Cancel the common factor of $x^{2}$ and $x$ .

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Step 4.3.3.1

Factor $x$ out of $2 x^{2}$ .

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

Step 4.3.3.2

Cancel the common factors.

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Step 4.3.3.2.1

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

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

Step 4.3.3.2.2

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

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

Step 4.3.3.2.3

Cancel the common factor.

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

Step 4.3.3.2.4

Rewrite the expression.

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

Step 4.3.3.2.5

Divide $2 x$ by $1$ .

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

Step 4.3.4

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

$y^{3} + y + C_{3} = \ln (| x |) + C_{4} + \int 2 x d x$

Step 4.3.5

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$y^{3} + y + C_{3} = \ln (| x |) + C_{4} + 2 \int x d x$

Step 4.3.6

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

$y^{3} + y + C_{3} = \ln (| x |) + C_{4} + 2 (\frac{1}{2} x^{2} + C_{5})$

Step 4.3.7

Simplify.

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Step 4.3.7.1

Simplify.

$y^{3} + y + C_{3} = \ln (| x |) + 2 (\frac{1}{2}) x^{2} + C_{6}$

Step 4.3.7.2

Simplify.

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Step 4.3.7.2.1

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

$y^{3} + y + C_{3} = \ln (| x |) + \frac{2}{2} x^{2} + C_{6}$

Step 4.3.7.2.2

Cancel the common factor of $2$ .

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Step 4.3.7.2.2.1

Cancel the common factor.

$y^{3} + y + C_{3} = \ln (| x |) + \frac{2}{2} x^{2} + C_{6}$

Step 4.3.7.2.2.2

Rewrite the expression.

$y^{3} + y + C_{3} = \ln (| x |) + 1 x^{2} + C_{6}$

Step 4.3.7.2.3

Multiply $x^{2}$ by $1$ .

$y^{3} + y + C_{3} = \ln (| x |) + x^{2} + C_{6}$

Step 4.3.8

Reorder terms.

$y^{3} + y + C_{3} = x^{2} + \ln (| x |) + C_{6}$

Step 4.4

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

$y^{3} + y = x^{2} + \ln (| x |) + C$