Calculus Examples

$\int_{1}^{9} (2 x + \frac{1}{x}) d x$

Step 1

Remove parentheses.

$\int_{1}^{9} 2 x + \frac{1}{x} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{1}^{9} 2 x d x + \int_{1}^{9} \frac{1}{x} d x$

Step 3

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int_{1}^{9} x d x + \int_{1}^{9} \frac{1}{x} d x$

Step 4

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

$2 (\frac{1}{2} x^{2}]_{1}^{9}) + \int_{1}^{9} \frac{1}{x} d x$

Step 5

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

$2 (\frac{x^{2}}{2}]_{1}^{9}) + \int_{1}^{9} \frac{1}{x} d x$

Step 6

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

$2 (\frac{x^{2}}{2}]_{1}^{9}) + \ln (| x |)]_{1}^{9}$

Step 7

Simplify the answer.

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Step 7.1

Substitute and simplify.

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Step 7.1.1

Evaluate $\frac{x^{2}}{2}$ at $9$ and at $1$ .

$2 ((\frac{9^{2}}{2}) - \frac{1^{2}}{2}) + \ln (| x |)]_{1}^{9}$

Step 7.1.2

Evaluate $\ln (| x |)$ at $9$ and at $1$ .

$2 (\frac{9^{2}}{2} - \frac{1^{2}}{2}) + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3

Simplify.

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Step 7.1.3.1

Raise $9$ to the power of $2$ .

$2 (\frac{81}{2} - \frac{1^{2}}{2}) + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.2

One to any power is one.

$2 (\frac{81}{2} - \frac{1}{2}) + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.3

Combine the numerators over the common denominator.

$2 \frac{81 - 1}{2} + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.4

Subtract $1$ from $81$ .

$2 (\frac{80}{2}) + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.5

Cancel the common factor of $80$ and $2$ .

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Step 7.1.3.5.1

Factor $2$ out of $80$ .

$2 \frac{2 \cdot 40}{2} + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.5.2

Cancel the common factors.

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Step 7.1.3.5.2.1

Factor $2$ out of $2$ .

$2 \frac{2 \cdot 40}{2 (1)} + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.5.2.2

Cancel the common factor.

$2 \frac{2 \cdot 40}{2 \cdot 1} + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.5.2.3

Rewrite the expression.

$2 (\frac{40}{1}) + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.5.2.4

Divide $40$ by $1$ .

$2 \cdot 40 + \ln (| 9 |) - \ln (| 1 |)$

Step 7.1.3.6

Multiply $2$ by $40$ .

$80 + \ln (| 9 |) - \ln (| 1 |)$

Step 7.2

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

$80 + \ln (\frac{| 9 |}{| 1 |})$

Step 7.3

Simplify.

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Step 7.3.1

The absolute value is the distance between a number and zero. The distance between $0$ and $9$ is $9$ .

$80 + \ln (\frac{9}{| 1 |})$

Step 7.3.2

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

$80 + \ln (\frac{9}{1})$

Step 7.3.3

Divide $9$ by $1$ .

$80 + \ln (9)$

Step 8

The result can be shown in multiple forms.

Exact Form:

$80 + \ln (9)$

Decimal Form:

$82.19722457 \dots$

Step 9