Calculus Examples

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

Step 1

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int {(3 x + 2)}^{- 2} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Let $u = 3 x + 2$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.1.1

Let $u = 3 x + 2$ . Find $\frac{d u}{d x}$ .

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Step 2.3.1.1.1

Differentiate $3 x + 2$ .

$\frac{d}{d x} [3 x + 2]$

Step 2.3.1.1.2

By the Sum Rule, the derivative of $3 x + 2$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$

Step 2.3.1.1.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 2.3.1.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 2.3.1.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$3 \cdot 1 + \frac{d}{d x} [2]$

Step 2.3.1.1.3.3

Multiply $3$ by $1$ .

$3 + \frac{d}{d x} [2]$

Step 2.3.1.1.4

Differentiate using the Constant Rule.

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Step 2.3.1.1.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$3 + 0$

Step 2.3.1.1.4.2

Add $3$ and $0$ .

$3$

Step 2.3.1.2

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

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

Combine $u^{- 2}$ and $\frac{1}{3}$ .

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

Step 2.3.2.2

Move $u^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.3.3

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

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

Step 2.3.4

Apply basic rules of exponents.

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Step 2.3.4.1

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

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

Step 2.3.4.2

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

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Step 2.3.4.2.1

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

$y + C_{1} = \frac{1}{3} \int u^{2 \cdot - 1} d u$

Step 2.3.4.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.5

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

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

Step 2.3.6

Simplify.

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Step 2.3.6.1

Rewrite $\frac{1}{3} (- u^{- 1} + C_{2})$ as $\frac{1}{3} (- \frac{1}{u}) + C_{2}$ .

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

Step 2.3.6.2

Multiply $\frac{1}{3}$ by $\frac{1}{u}$ .

$y + C_{1} = - \frac{1}{3 u} + C_{2}$

Step 2.3.7

Replace all occurrences of $u$ with $3 x + 2$ .

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

Step 2.4

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

$y = - \frac{1}{3 (3 x + 2)} + K$