Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

Let $u_{2} = \ln (\frac{1}{x})$ . Find $\frac{d u_{2}}{d x}$ .

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Step 2.1.1

Differentiate $\ln (\frac{1}{x})$ .

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

Step 2.1.2

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 $g (x) = \frac{1}{x}$ .

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.3

Multiply by the reciprocal of the fraction to divide by $\frac{1}{x}$ .

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

Step 2.1.4

Simplify the expression.

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Step 2.1.4.1

Multiply $x$ by $1$ .

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

Step 2.1.4.2

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

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

Step 2.1.5

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

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

Step 2.1.6

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

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

Step 2.1.7

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

$- x^{1 - 2}$

Step 2.1.8

Subtract $2$ from $1$ .

$- x^{- 1}$

Step 2.1.9

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{x}$

Step 2.2

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

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

Step 3

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

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

Step 4

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 5

Rewrite $- (\frac{1}{2} {u_{2}}^{2} + C)$ as $- \frac{1}{2} {u_{2}}^{2} + C$ .

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

Step 6

Replace all occurrences of $u_{2}$ with $\ln (\frac{1}{x})$ .

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