Calculus Examples

$\int_{1}^{2} \frac{x - 4}{x^{2}} d x$

Step 1

This integral could not be completed using u-substitution. Mathway will use another method.

Step 2

Apply basic rules of exponents.

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Step 2.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

$\int_{1}^{2} (x - 4) {(x^{2})}^{- 1} d x$

Step 2.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 2.2.1

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

$\int_{1}^{2} (x - 4) x^{2 \cdot - 1} d x$

Step 2.2.2

Multiply $2$ by $- 1$ .

$\int_{1}^{2} (x - 4) x^{- 2} d x$

Step 3

Multiply $(x - 4) x^{- 2}$ .

$\int_{1}^{2} x \cdot x^{- 2} - 4 x^{- 2} d x$

Step 4

Multiply $x$ by $x^{- 2}$ by adding the exponents.

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Step 4.1

Multiply $x$ by $x^{- 2}$ .

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Step 4.1.1

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

$\int_{1}^{2} x^{1} x^{- 2} - 4 x^{- 2} d x$

Step 4.1.2

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

$\int_{1}^{2} x^{1 - 2} - 4 x^{- 2} d x$

Step 4.2

Subtract $2$ from $1$ .

$\int_{1}^{2} x^{- 1} - 4 x^{- 2} d x$

Step 5

Split the single integral into multiple integrals.

$\int_{1}^{2} x^{- 1} d x + \int_{1}^{2} - 4 x^{- 2} d x$

Step 6

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

$\ln (| x |)]_{1}^{2} + \int_{1}^{2} - 4 x^{- 2} d x$

Step 7

Since $- 4$ is constant with respect to $x$ , move $- 4$ out of the integral.

$\ln (| x |)]_{1}^{2} - 4 \int_{1}^{2} x^{- 2} d x$

Step 8

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

$\ln (| x |)]_{1}^{2} - 4 (- x^{- 1}]_{1}^{2})$

Step 9

Simplify the answer.

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Step 9.1

Substitute and simplify.

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Step 9.1.1

Evaluate $\ln (| x |)$ at $2$ and at $1$ .

$(\ln (| 2 |)) - \ln (| 1 |) - 4 (- x^{- 1}]_{1}^{2})$

Step 9.1.2

Evaluate $- x^{- 1}$ at $2$ and at $1$ .

$\ln (| 2 |) - \ln (| 1 |) - 4 (- 2^{- 1} + 1^{- 1})$

Step 9.1.3

Simplify.

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Step 9.1.3.1

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

$\ln (| 2 |) - \ln (| 1 |) - 4 (- \frac{1}{2} + 1^{- 1})$

Step 9.1.3.2

One to any power is one.

$\ln (| 2 |) - \ln (| 1 |) - 4 (- \frac{1}{2} + 1)$

Step 9.1.3.3

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

$\ln (| 2 |) - \ln (| 1 |) - 4 (- \frac{1}{2} + \frac{2}{2})$

Step 9.1.3.4

Combine the numerators over the common denominator.

$\ln (| 2 |) - \ln (| 1 |) - 4 \frac{- 1 + 2}{2}$

Step 9.1.3.5

Add $- 1$ and $2$ .

$\ln (| 2 |) - \ln (| 1 |) - 4 (\frac{1}{2})$

Step 9.1.3.6

Combine $- 4$ and $\frac{1}{2}$ .

$\ln (| 2 |) - \ln (| 1 |) + \frac{- 4}{2}$

Step 9.1.3.7

Cancel the common factor of $- 4$ and $2$ .

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Step 9.1.3.7.1

Factor $2$ out of $- 4$ .

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

Step 9.1.3.7.2

Cancel the common factors.

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Step 9.1.3.7.2.1

Factor $2$ out of $2$ .

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

Step 9.1.3.7.2.2

Cancel the common factor.

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

Step 9.1.3.7.2.3

Rewrite the expression.

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

Step 9.1.3.7.2.4

Divide $- 2$ by $1$ .

$\ln (| 2 |) - \ln (| 1 |) - 2$

Step 9.2

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

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

Step 9.3

Simplify.

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Step 9.3.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Step 9.3.2

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

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

Step 9.3.3

Divide $2$ by $1$ .

$\ln (2) - 2$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\ln (2) - 2$

Decimal Form:

$- 1.30685281 \dots$

Step 11