Calculus Examples

$\frac{d y}{d x} = 4 x + \frac{4 x}{\sqrt{16 - x^{2}}}$

Step 1

Rewrite the equation.

$d y = (4 x + \frac{4 x}{\sqrt{16 - x^{2}}}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 4 x + \frac{4 x}{\sqrt{16 - x^{2}}} d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 4 x + \frac{4 x}{\sqrt{16 - x^{2}}} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 4 x d x + \int \frac{4 x}{\sqrt{16 - x^{2}}} d x$

Step 2.3.2

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

$y + C_{1} = 4 \int x d x + \int \frac{4 x}{\sqrt{16 - x^{2}}} d x$

Step 2.3.3

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

$y + C_{1} = 4 (\frac{1}{2} x^{2} + C_{2}) + \int \frac{4 x}{\sqrt{16 - x^{2}}} d x$

Step 2.3.4

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

$y + C_{1} = 4 (\frac{1}{2} x^{2} + C_{2}) + 4 \int \frac{x}{\sqrt{16 - x^{2}}} d x$

Step 2.3.5

Let $u = 16 - x^{2}$ . 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 2.3.5.1

Let $u = 16 - x^{2}$ . Find $\frac{d u}{d x}$ .

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Step 2.3.5.1.1

Differentiate $16 - x^{2}$ .

$\frac{d}{d x} [16 - x^{2}]$

Step 2.3.5.1.2

Differentiate.

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Step 2.3.5.1.2.1

By the Sum Rule, the derivative of $16 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [16] + \frac{d}{d x} [- x^{2}]$ .

$\frac{d}{d x} [16] + \frac{d}{d x} [- x^{2}]$

Step 2.3.5.1.2.2

Since $16$ is constant with respect to $x$ , the derivative of $16$ with respect to $x$ is $0$ .

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

Step 2.3.5.1.3

Evaluate $\frac{d}{d x} [- x^{2}]$ .

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Step 2.3.5.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 2.3.5.1.3.2

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 2.3.5.1.3.3

Multiply $2$ by $- 1$ .

$0 - 2 x$

Step 2.3.5.1.4

Subtract $2 x$ from $0$ .

$- 2 x$

Step 2.3.5.2

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

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

Step 2.3.6

Simplify.

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Step 2.3.6.1

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

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

Step 2.3.6.2

Move the negative in front of the fraction.

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

Step 2.3.6.3

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

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

Step 2.3.6.4

Move $2$ to the left of $\sqrt{u}$ .

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

Step 2.3.7

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

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

Step 2.3.8

Multiply $- 1$ by $4$ .

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

Step 2.3.9

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

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

Step 2.3.10

Simplify the expression.

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Step 2.3.10.1

Simplify.

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Step 2.3.10.1.1

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

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

Step 2.3.10.1.2

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

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Step 2.3.10.1.2.1

Factor $2$ out of $- 4$ .

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

Step 2.3.10.1.2.2

Cancel the common factors.

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Step 2.3.10.1.2.2.1

Factor $2$ out of $2$ .

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

Step 2.3.10.1.2.2.2

Cancel the common factor.

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

Step 2.3.10.1.2.2.3

Rewrite the expression.

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

Step 2.3.10.1.2.2.4

Divide $- 2$ by $1$ .

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

Step 2.3.10.2

Apply basic rules of exponents.

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Step 2.3.10.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u}$ as $u^{\frac{1}{2}}$ .

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

Step 2.3.10.2.2

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

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

Step 2.3.10.2.3

Multiply the exponents in ${(u^{\frac{1}{2}})}^{- 1}$ .

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Step 2.3.10.2.3.1

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

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

Step 2.3.10.2.3.2

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

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

Step 2.3.10.2.3.3

Move the negative in front of the fraction.

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

Step 2.3.11

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

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

Step 2.3.12

Simplify.

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Step 2.3.12.1

Simplify.

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

Step 2.3.12.2

Multiply $2$ by $- 2$ .

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

Step 2.3.13

Replace all occurrences of $u$ with $16 - x^{2}$ .

$y + C_{1} = 2 x^{2} - 4 {(16 - x^{2})}^{\frac{1}{2}} + C_{4}$

Step 2.4

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

$y = 2 x^{2} - 4 {(16 - x^{2})}^{\frac{1}{2}} + K$