Calculus Examples

$\frac{\ln (x)}{x^{5}}$

Step 1

Write $\frac{\ln (x)}{x^{5}}$ as a function.

$f (x) = \frac{\ln (x)}{x^{5}}$

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.

$F (x) = \int \frac{\ln (x)}{x^{5}} d x$

Step 4

Apply basic rules of exponents.

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Step 4.1

Move $x^{5}$ out of the denominator by raising it to the $- 1$ power.

$\int \ln (x) {(x^{5})}^{- 1} d x$

Step 4.2

Multiply the exponents in ${(x^{5})}^{- 1}$ .

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Step 4.2.1

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

$\int \ln (x) x^{5 \cdot - 1} d x$

Step 4.2.2

Multiply $5$ by $- 1$ .

$\int \ln (x) x^{- 5} d x$

Step 5

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = x^{- 5}$ .

$\ln (x) (- \frac{1}{4 x^{4}}) - \int - \frac{1}{4 x^{4}} \cdot \frac{1}{x} d x$

Step 6

Simplify.

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Step 6.1

Combine $\ln (x)$ and $\frac{1}{4 x^{4}}$ .

$- \frac{\ln (x)}{4 x^{4}} - \int - \frac{1}{4 x^{4}} \cdot \frac{1}{x} d x$

Step 6.2

Multiply $\frac{1}{x}$ by $\frac{1}{4 x^{4}}$ .

$- \frac{\ln (x)}{4 x^{4}} - \int - \frac{1}{x (4 x^{4})} d x$

Step 6.3

Raise $x$ to the power of $1$ .

$- \frac{\ln (x)}{4 x^{4}} - \int - \frac{1}{4 (x^{1} x^{4})} d x$

Step 6.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.5

Add $1$ and $4$ .

$- \frac{\ln (x)}{4 x^{4}} - \int - \frac{1}{4 x^{5}} d x$

Step 7

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

$- \frac{\ln (x)}{4 x^{4}} - - \int \frac{1}{4 x^{5}} d x$

Step 8

Simplify.

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Step 8.1

Multiply $- 1$ by $- 1$ .

$- \frac{\ln (x)}{4 x^{4}} + 1 \int \frac{1}{4 x^{5}} d x$

Step 8.2

Multiply $\int \frac{1}{4 x^{5}} d x$ by $1$ .

$- \frac{\ln (x)}{4 x^{4}} + \int \frac{1}{4 x^{5}} d x$

Step 9

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$- \frac{\ln (x)}{4 x^{4}} + \frac{1}{4} \int \frac{1}{x^{5}} d x$

Step 10

Apply basic rules of exponents.

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Step 10.1

Move $x^{5}$ out of the denominator by raising it to the $- 1$ power.

$- \frac{\ln (x)}{4 x^{4}} + \frac{1}{4} \int {(x^{5})}^{- 1} d x$

Step 10.2

Multiply the exponents in ${(x^{5})}^{- 1}$ .

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Step 10.2.1

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

$- \frac{\ln (x)}{4 x^{4}} + \frac{1}{4} \int x^{5 \cdot - 1} d x$

Step 10.2.2

Multiply $5$ by $- 1$ .

$- \frac{\ln (x)}{4 x^{4}} + \frac{1}{4} \int x^{- 5} d x$

Step 11

By the Power Rule, the integral of $x^{- 5}$ with respect to $x$ is $- \frac{1}{4} x^{- 4}$ .

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

Step 12

Simplify the answer.

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Step 12.1

Rewrite $- \frac{\ln (x)}{4 x^{4}} + \frac{1}{4} (- \frac{1}{4} x^{- 4} + C)$ as $- \frac{\ln (x)}{4 x^{4}} + \frac{1}{4} (- \frac{1}{4} \cdot \frac{1}{x^{4}}) + C$ .

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

Step 12.2

Simplify.

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Step 12.2.1

Multiply $\frac{1}{x^{4}}$ by $\frac{1}{4}$ .

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

Step 12.2.2

Move $4$ to the left of $x^{4}$ .

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

Step 12.2.3

Multiply $\frac{1}{4}$ by $\frac{1}{4 x^{4}}$ .

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

Step 12.2.4

Multiply $4$ by $4$ .

$- \frac{\ln (x)}{4 x^{4}} - \frac{1}{16 x^{4}} + C$

Step 13

The answer is the antiderivative of the function $f (x) = \frac{\ln (x)}{x^{5}}$ .

$F (x) =$ $- \frac{\ln (x)}{4 x^{4}} - \frac{1}{16 x^{4}} + C$