Calculus Examples

\frac{2}{x^{3}} - \frac{4}{x^{2}}

$\frac{2}{x^{3}} - \frac{4}{x^{2}}$

Step 1

Write

\frac{2}{x^{3}} - \frac{4}{x^{2}}

$\frac{2}{x^{3}} - \frac{4}{x^{2}}$ as a function.

f (x) = \frac{2}{x^{3}} - \frac{4}{x^{2}}

$f (x) = \frac{2}{x^{3}} - \frac{4}{x^{2}}$

Step 2

The function

F (x)

$F (x)$ can be found by finding the indefinite integral of the derivative

f (x)

$f (x)$ .

F (x) = \int f (x) d x

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

F (x) = \int \frac{2}{x^{3}} - \frac{4}{x^{2}} d x

$F (x) = \int \frac{2}{x^{3}} - \frac{4}{x^{2}} d x$

Step 4

Split the single integral into multiple integrals.

\int \frac{2}{x^{3}} d x + \int - \frac{4}{x^{2}} d x

$\int \frac{2}{x^{3}} d x + \int - \frac{4}{x^{2}} d x$

Step 5

Since

2

$2$ is constant with respect to

x

$x$ , move

2

$2$ out of the integral.

2 \int \frac{1}{x^{3}} d x + \int - \frac{4}{x^{2}} d x

$2 \int \frac{1}{x^{3}} d x + \int - \frac{4}{x^{2}} d x$

Step 6

Apply basic rules of exponents.

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Step 6.1

Move

x^{3}

$x^{3}$ out of the denominator by raising it to the

- 1

$- 1$ power.

2 \int {(x^{3})}^{- 1} d x + \int - \frac{4}{x^{2}} d x

$2 \int {(x^{3})}^{- 1} d x + \int - \frac{4}{x^{2}} d x$

Step 6.2

Multiply the exponents in

{(x^{3})}^{- 1}

${(x^{3})}^{- 1}$ .

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Step 6.2.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

2 \int x^{3 \cdot - 1} d x + \int - \frac{4}{x^{2}} d x

$2 \int x^{3 \cdot - 1} d x + \int - \frac{4}{x^{2}} d x$

Step 6.2.2

Multiply

3

$3$ by

- 1

$- 1$ .

2 \int x^{- 3} d x + \int - \frac{4}{x^{2}} d x

$2 \int x^{- 3} d x + \int - \frac{4}{x^{2}} d x$

2 \int x^{- 3} d x + \int - \frac{4}{x^{2}} d x

$2 \int x^{- 3} d x + \int - \frac{4}{x^{2}} d x$

2 \int x^{- 3} d x + \int - \frac{4}{x^{2}} d x

$2 \int x^{- 3} d x + \int - \frac{4}{x^{2}} d x$

Step 7

By the Power Rule, the integral of

x^{- 3}

$x^{- 3}$ with respect to

x

$x$ is

- \frac{1}{2} x^{- 2}

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

2 (- \frac{1}{2} x^{- 2} + C) + \int - \frac{4}{x^{2}} d x

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

Step 8

Simplify.

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Step 8.1

Combine

x^{- 2}

$x^{- 2}$ and

\frac{1}{2}

$\frac{1}{2}$ .

2 (- \frac{x^{- 2}}{2} + C) + \int - \frac{4}{x^{2}} d x

$2 (- \frac{x^{- 2}}{2} + C) + \int - \frac{4}{x^{2}} d x$

Step 8.2

Move

x^{- 2}

$x^{- 2}$ to the denominator using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

2 (- \frac{1}{2 x^{2}} + C) + \int - \frac{4}{x^{2}} d x

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

2 (- \frac{1}{2 x^{2}} + C) + \int - \frac{4}{x^{2}} d x

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

Step 9

Since

- 1

$- 1$ is constant with respect to

x

$x$ , move

- 1

$- 1$ out of the integral.

2 (- \frac{1}{2 x^{2}} + C) - \int \frac{4}{x^{2}} d x

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

Step 10

Since

4

$4$ is constant with respect to

x

$x$ , move

4

$4$ out of the integral.

2 (- \frac{1}{2 x^{2}} + C) - (4 \int \frac{1}{x^{2}} d x)

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

Step 11

Simplify the expression.

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Step 11.1

Multiply

4

$4$ by

- 1

$- 1$ .

2 (- \frac{1}{2 x^{2}} + C) - 4 \int \frac{1}{x^{2}} d x

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

Step 11.2

Move

x^{2}

$x^{2}$ out of the denominator by raising it to the

- 1

$- 1$ power.

2 (- \frac{1}{2 x^{2}} + C) - 4 \int {(x^{2})}^{- 1} d x

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

Step 11.3

Multiply the exponents in

{(x^{2})}^{- 1}

${(x^{2})}^{- 1}$ .

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Step 11.3.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

2 (- \frac{1}{2 x^{2}} + C) - 4 \int x^{2 \cdot - 1} d x

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

Step 11.3.2

Multiply

2

$2$ by

- 1

$- 1$ .

2 (- \frac{1}{2 x^{2}} + C) - 4 \int x^{- 2} d x

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

2 (- \frac{1}{2 x^{2}} + C) - 4 \int x^{- 2} d x

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

2 (- \frac{1}{2 x^{2}} + C) - 4 \int x^{- 2} d x

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

Step 12

By the Power Rule, the integral of

x^{- 2}

$x^{- 2}$ with respect to

x

$x$ is

- x^{- 1}

$- x^{- 1}$ .

2 (- \frac{1}{2 x^{2}} + C) - 4 (- x^{- 1} + C)

$2 (- \frac{1}{2 x^{2}} + C) - 4 (- x^{- 1} + C)$

Step 13

Simplify.

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Step 13.1

Simplify.

\frac{- 1}{x^{2}} - 4 (- x^{- 1}) + C

$\frac{- 1}{x^{2}} - 4 (- x^{- 1}) + C$

Step 13.2

Simplify.

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Step 13.2.1

Move the negative in front of the fraction.

- \frac{1}{x^{2}} - 4 (- x^{- 1}) + C

$- \frac{1}{x^{2}} - 4 (- x^{- 1}) + C$

Step 13.2.2

Multiply

- 1

$- 1$ by

- 4

$- 4$ .

- \frac{1}{x^{2}} + 4 x^{- 1} + C

$- \frac{1}{x^{2}} + 4 x^{- 1} + C$

- \frac{1}{x^{2}} + 4 x^{- 1} + C

$- \frac{1}{x^{2}} + 4 x^{- 1} + C$

- \frac{1}{x^{2}} + 4 x^{- 1} + C

$- \frac{1}{x^{2}} + 4 x^{- 1} + C$

Step 14

The answer is the antiderivative of the function

f (x) = \frac{2}{x^{3}} - \frac{4}{x^{2}}

$f (x) = \frac{2}{x^{3}} - \frac{4}{x^{2}}$ .

F (x) =

$F (x) =$

- \frac{1}{x^{2}} + 4 x^{- 1} + C

$- \frac{1}{x^{2}} + 4 x^{- 1} + C$