Calculus Examples

${(\frac{1}{x})}^{2}$

Step 1

Write ${(\frac{1}{x})}^{2}$ as a function.

$f (x) = {(\frac{1}{x})}^{2}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

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

Step 3

Set up the integral to solve.

$F (x) = \int {(\frac{1}{x})}^{2} d x$

Step 4

Simplify the expression.

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Step 4.1

Apply the product rule to $\frac{1}{x}$ .

$\int \frac{1^{2}}{x^{2}} d x$

Step 4.2

One to any power is one.

$\int \frac{1}{x^{2}} d x$

Step 4.3

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

$\int {(x^{2})}^{- 1} d x$

Step 4.4

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

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Step 4.4.1

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

$\int x^{2 \cdot - 1} d x$

Step 4.4.2

Multiply $2$ by $- 1$ .

$\int x^{- 2} d x$

Step 5

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

$- x^{- 1} + C$

Step 6

Rewrite $- x^{- 1} + C$ as $- \frac{1}{x} + C$ .

$- \frac{1}{x} + C$

Step 7

The answer is the antiderivative of the function $f (x) = {(\frac{1}{x})}^{2}$ .

$F (x) =$ $- \frac{1}{x} + C$