Calculus Examples

$\int_{1}^{e} \frac{1 - \ln (x)}{x} d x$

Step 1

Split the fraction into multiple fractions.

$\int_{1}^{e} \frac{1}{x} + \frac{- \ln (x)}{x} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{1}^{e} \frac{1}{x} d x + \int_{1}^{e} \frac{- \ln (x)}{x} d x$

Step 3

Move the negative in front of the fraction.

$\int_{1}^{e} \frac{1}{x} d x + \int_{1}^{e} - \frac{\ln (x)}{x} d x$

Step 4

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

$\ln (| x |)]_{1}^{e} + \int_{1}^{e} - \frac{\ln (x)}{x} d x$

Step 5

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

$\ln (| x |)]_{1}^{e} - \int_{1}^{e} \frac{\ln (x)}{x} d x$

Step 6

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 6.1

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

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Step 6.1.1

Differentiate $\ln (x)$ .

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

Step 6.1.2

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

$\frac{1}{x}$

Step 6.2

Substitute the lower limit in for $x$ in $u = \ln (x)$ .

$u_{lower} = \ln (1)$

Step 6.3

The natural logarithm of $1$ is $0$ .

$u_{lower} = 0$

Step 6.4

Substitute the upper limit in for $x$ in $u = \ln (x)$ .

$u_{upper} = \ln (e)$

Step 6.5

The natural logarithm of $e$ is $1$ .

$u_{upper} = 1$

Step 6.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = 1$

Step 6.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$\ln (| x |)]_{1}^{e} - \int_{0}^{1} u d u$

Step 7

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

$\ln (| x |)]_{1}^{e} - (\frac{1}{2} u^{2}]_{0}^{1})$

Step 8

Combine $\frac{1}{2}$ and $u^{2}$ .

$\ln (| x |)]_{1}^{e} - (\frac{u^{2}}{2}]_{0}^{1})$

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $\ln (| x |)$ at $e$ and at $1$ .

$(\ln (| e |)) - \ln (| 1 |) - (\frac{u^{2}}{2}]_{0}^{1})$

Step 9.2

Evaluate $\frac{u^{2}}{2}$ at $1$ and at $0$ .

$(\ln (| e |)) - \ln (| 1 |) - ((\frac{1^{2}}{2}) - \frac{0^{2}}{2})$

Step 9.3

Simplify.

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Step 9.3.1

One to any power is one.

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - \frac{0^{2}}{2})$

Step 9.3.2

Raising $0$ to any positive power yields $0$ .

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - \frac{0}{2})$

Step 9.3.3

Cancel the common factor of $0$ and $2$ .

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Step 9.3.3.1

Factor $2$ out of $0$ .

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - \frac{2 (0)}{2})$

Step 9.3.3.2

Cancel the common factors.

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Step 9.3.3.2.1

Factor $2$ out of $2$ .

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - \frac{2 \cdot 0}{2 \cdot 1})$

Step 9.3.3.2.2

Cancel the common factor.

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - \frac{2 \cdot 0}{2 \cdot 1})$

Step 9.3.3.2.3

Rewrite the expression.

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - \frac{0}{1})$

Step 9.3.3.2.4

Divide $0$ by $1$ .

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} - 0)$

Step 9.3.4

Multiply $- 1$ by $0$ .

$\ln (| e |) - \ln (| 1 |) - (\frac{1}{2} + 0)$

Step 9.3.5

Add $\frac{1}{2}$ and $0$ .

$\ln (| e |) - \ln (| 1 |) - \frac{1}{2}$

Step 10

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| e |}{| 1 |}) - \frac{1}{2}$

Step 11

Simplify.

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Step 11.1

$e$ is approximately $2.71828182$ which is positive so remove the absolute value

$\ln (\frac{e}{| 1 |}) - \frac{1}{2}$

Step 11.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\ln (\frac{e}{1}) - \frac{1}{2}$

Step 11.3

Divide $e$ by $1$ .

$\ln (e) - \frac{1}{2}$

Step 12

Simplify.

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Step 12.1

The natural logarithm of $e$ is $1$ .

$1 - \frac{1}{2}$

Step 12.2

Write $1$ as a fraction with a common denominator.

$\frac{2}{2} - \frac{1}{2}$

Step 12.3

Combine the numerators over the common denominator.

$\frac{2 - 1}{2}$

Step 12.4

Subtract $1$ from $2$ .

$\frac{1}{2}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{2}$

Decimal Form:

$0.5$