Calculus Examples

$\int \frac{\ln (8 x)}{x} d x$

Step 1

Let $u_{2} = \ln (8 x)$ . Then $d u_{2} = \frac{1}{x} d x$ , so $x d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 1.1

Let $u_{2} = \ln (8 x)$ . Find $\frac{d u_{2}}{d x}$ .

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Step 1.1.1

Differentiate $\ln (8 x)$ .

$\frac{d}{d x} [\ln (8 x)]$

Step 1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 8 x$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u_{1}$ as $8 x$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [8 x]$

Step 1.1.2.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [8 x]$

Step 1.1.2.3

Replace all occurrences of $u_{1}$ with $8 x$ .

$\frac{1}{8 x} \frac{d}{d x} [8 x]$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$\frac{1}{8 x} (8 \frac{d}{d x} [x])$

Step 1.1.3.2

Simplify terms.

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Step 1.1.3.2.1

Combine $8$ and $\frac{1}{8 x}$ .

$\frac{8}{8 x} \frac{d}{d x} [x]$

Step 1.1.3.2.2

Cancel the common factor of $8$ .

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Step 1.1.3.2.2.1

Cancel the common factor.

$\frac{8}{8 x} \frac{d}{d x} [x]$

Step 1.1.3.2.2.2

Rewrite the expression.

$\frac{1}{x} \frac{d}{d x} [x]$

Step 1.1.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{x} \cdot 1$

Step 1.1.3.4

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

$\frac{1}{x}$

Step 1.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\int u_{2} d u_{2}$

Step 2

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

$\frac{1}{2} {u_{2}}^{2} + C$

Step 3

Replace all occurrences of $u_{2}$ with $\ln (8 x)$ .

$\frac{1}{2} \ln^{2} (8 x) + C$