Calculus Examples

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

Step 1

Let $u = \ln (x)$ . Then $d u = \frac{1}{x} d x$ , so $x d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = \ln (x)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $\ln (x)$ .

$\frac{d}{d x} [\ln (x)]$

Step 1.1.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\frac{1}{x}$

Step 1.2

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

$\int \frac{1}{u^{4}} d u$

Step 2

Apply basic rules of exponents.

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Step 2.1

Move $u^{4}$ out of the denominator by raising it to the $- 1$ power.

$\int {(u^{4})}^{- 1} d u$

Step 2.2

Multiply the exponents in ${(u^{4})}^{- 1}$ .

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Step 2.2.1

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

$\int u^{4 \cdot - 1} d u$

Step 2.2.2

Multiply $4$ by $- 1$ .

$\int u^{- 4} d u$

Step 3

By the Power Rule, the integral of $u^{- 4}$ with respect to $u$ is $- \frac{1}{3} u^{- 3}$ .

$- \frac{1}{3} u^{- 3} + C$

Step 4

Simplify.

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Step 4.1

Rewrite $- \frac{1}{3} u^{- 3} + C$ as $- \frac{1}{3} \cdot \frac{1}{u^{3}} + C$ .

$- \frac{1}{3} \cdot \frac{1}{u^{3}} + C$

Step 4.2

Simplify.

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Step 4.2.1

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

$- \frac{1}{u^{3} \cdot 3} + C$

Step 4.2.2

Move $3$ to the left of $u^{3}$ .

$- \frac{1}{3 u^{3}} + C$

Step 5

Replace all occurrences of $u$ with $\ln (x)$ .

$- \frac{1}{3 \ln^{3} (x)} + C$