Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Divide each term in $x \frac{d y}{d x} = y^{2} + 1$ by $x$ and simplify.

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Step 1.1.1

Divide each term in $x \frac{d y}{d x} = y^{2} + 1$ by $x$ .

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

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $x$ .

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Step 1.1.2.1.1

Cancel the common factor.

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

Step 1.1.2.1.2

Divide $\frac{d y}{d x}$ by $1$ .

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

Step 1.2

Combine the numerators over the common denominator.

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

Step 1.3

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

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

Step 1.4

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

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Reorder $y^{2}$ and $1$ .

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

Step 2.2.1.2

Rewrite $1$ as $1^{2}$ .

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

Step 2.2.2

The integral of $\frac{1}{1^{2} + y^{2}}$ with respect to $y$ is $\arctan (y) + C_{1}$ .

$\arctan (y) + C_{1} = \int \frac{1}{x} d x$

Step 2.3

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

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

Step 2.4

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

$\arctan (y) = \ln (| x |) + C$

Step 3

Take the inverse arctangent of both sides of the equation to extract $y$ from inside the arctangent.

$y = \tan (\ln (| x |) + C)$