Calculus Examples

$(4 + x^{2}) d y + 4 d x = 0$

Step 1

Subtract $4 d x$ from both sides of the equation.

$(4 + x^{2}) d y = - 4 d x$

Step 2

Multiply both sides by $\frac{1}{4 + x^{2}}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $4 + x^{2}$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

$1 d y = \frac{1}{4 + x^{2}} \cdot - 4 d x$

Step 3.2

Combine $\frac{1}{4 + x^{2}}$ and $- 4$ .

$1 d y = \frac{- 4}{4 + x^{2}} d x$

Step 3.3

Move the negative in front of the fraction.

$1 d y = - \frac{4}{4 + x^{2}} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int d y = \int - \frac{4}{4 + x^{2}} d x$

Step 4.2

Apply the constant rule.

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 4.3.3

Write the expression using exponents.

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Step 4.3.3.1

Multiply $4$ by $- 1$ .

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

Step 4.3.3.2

Rewrite $4$ as $2^{2}$ .

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

Step 4.3.4

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

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

Step 4.3.5

Simplify the answer.

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Step 4.3.5.1

Combine $\frac{1}{2}$ and $\arctan (\frac{x}{2})$ .

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

Step 4.3.5.2

Rewrite $- 4 (\frac{\arctan (\frac{x}{2})}{2} + C_{2})$ as $- 4 (\frac{1}{2}) \arctan (\frac{1}{2} x) + C_{2}$ .

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

Step 4.3.5.3

Simplify.

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Step 4.3.5.3.1

Combine $- 4$ and $\frac{1}{2}$ .

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

Step 4.3.5.3.2

Cancel the common factor of $- 4$ and $2$ .

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Step 4.3.5.3.2.1

Factor $2$ out of $- 4$ .

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

Step 4.3.5.3.2.2

Cancel the common factors.

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Step 4.3.5.3.2.2.1

Factor $2$ out of $2$ .

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

Step 4.3.5.3.2.2.2

Cancel the common factor.

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

Step 4.3.5.3.2.2.3

Rewrite the expression.

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

Step 4.3.5.3.2.2.4

Divide $- 2$ by $1$ .

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

Step 4.3.5.3.3

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

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

Step 4.3.5.4

Reorder terms.

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

Step 4.4

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

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