Calculus Examples

$\frac{d y}{d x} = - \frac{1}{x^{3}}$

Step 1

Rewrite the equation.

$d y = - \frac{1}{x^{3}} d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int - \frac{1}{x^{3}} d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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

Step 2.3.2

Apply basic rules of exponents.

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Step 2.3.2.1

Move $x^{3}$ out of the denominator by raising it to the $- 1$ power.

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

Step 2.3.2.2

Multiply the exponents in ${(x^{3})}^{- 1}$ .

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Step 2.3.2.2.1

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

$y + C_{1} = - \int x^{3 \cdot - 1} d x$

Step 2.3.2.2.2

Multiply $3$ by $- 1$ .

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

Step 2.3.3

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

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

Step 2.3.4

Simplify the answer.

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Step 2.3.4.1

Simplify.

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Step 2.3.4.1.1

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

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

Step 2.3.4.1.2

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

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

Step 2.3.4.2

Simplify.

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

Step 2.3.4.3

Simplify.

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Step 2.3.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.4.3.2

Multiply $\frac{1}{2 x^{2}}$ by $1$ .

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

Step 2.4

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

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