Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

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 $1$ .

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

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Apply basic rules of exponents.

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Step 2.3.1.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.1.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 2.3.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.2

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

$\arctan (y) + C_{1} = - x^{- 1} + C_{2}$

Step 2.3.3

Rewrite $- x^{- 1} + C_{2}$ as $- \frac{1}{x} + C_{2}$ .

$\arctan (y) + C_{1} = - \frac{1}{x} + C_{2}$

Step 2.4

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

$\arctan (y) = - \frac{1}{x} + K$

Step 3

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

$y = \tan (- \frac{1}{x} + K)$