Calculus Examples

$\int_{0}^{1} (\frac{1}{x + 1} + \frac{1}{{(x + 4)}^{2}}) d x$

Step 1

Remove parentheses.

$\int_{0}^{1} \frac{1}{x + 1} + \frac{1}{{(x + 4)}^{2}} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{0}^{1} \frac{1}{x + 1} d x + \int_{0}^{1} \frac{1}{{(x + 4)}^{2}} d x$

Step 3

Let $u_{1} = x + 1$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 3.1

Let $u_{1} = x + 1$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $x + 1$ .

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

Step 3.1.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

Step 3.1.3

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

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

Step 3.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 3.1.5

Add $1$ and $0$ .

$1$

Step 3.2

Substitute the lower limit in for $x$ in $u_{1} = x + 1$ .

$u_{lower} = 0 + 1$

Step 3.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 3.4

Substitute the upper limit in for $x$ in $u_{1} = x + 1$ .

$u_{upper} = 1 + 1$

Step 3.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 3.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 3.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$\int_{1}^{2} \frac{1}{u_{1}} d u_{1} + \int_{0}^{1} \frac{1}{{(x + 4)}^{2}} d x$

Step 4

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

$\ln (| u_{1} |)]_{1}^{2} + \int_{0}^{1} \frac{1}{{(x + 4)}^{2}} d x$

Step 5

Let $u_{2} = x + 4$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 5.1

Let $u_{2} = x + 4$ . Find $\frac{d u_{2}}{d x}$ .

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Step 5.1.1

Differentiate $x + 4$ .

$\frac{d}{d x} [x + 4]$

Step 5.1.2

By the Sum Rule, the derivative of $x + 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [4]$

Step 5.1.3

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

$1 + \frac{d}{d x} [4]$

Step 5.1.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Substitute the lower limit in for $x$ in $u_{2} = x + 4$ .

$u_{lower} = 0 + 4$

Step 5.3

Add $0$ and $4$ .

$u_{lower} = 4$

Step 5.4

Substitute the upper limit in for $x$ in $u_{2} = x + 4$ .

$u_{upper} = 1 + 4$

Step 5.5

Add $1$ and $4$ .

$u_{upper} = 5$

Step 5.6

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

$u_{lower} = 4$

$u_{upper} = 5$

Step 5.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$\ln (| u_{1} |)]_{1}^{2} + \int_{4}^{5} \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 6

Apply basic rules of exponents.

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Step 6.1

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

$\ln (| u_{1} |)]_{1}^{2} + \int_{4}^{5} {({u_{2}}^{2})}^{- 1} d u_{2}$

Step 6.2

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

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Step 6.2.1

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

$\ln (| u_{1} |)]_{1}^{2} + \int_{4}^{5} {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 6.2.2

Multiply $2$ by $- 1$ .

$\ln (| u_{1} |)]_{1}^{2} + \int_{4}^{5} {u_{2}}^{- 2} d u_{2}$

Step 7

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

$\ln (| u_{1} |)]_{1}^{2} + - {u_{2}}^{- 1}]_{4}^{5}$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $\ln (| u_{1} |)$ at $2$ and at $1$ .

$(\ln (| 2 |)) - \ln (| 1 |) + - {u_{2}}^{- 1}]_{4}^{5}$

Step 8.2

Evaluate $- {u_{2}}^{- 1}$ at $5$ and at $4$ .

$(\ln (| 2 |)) - \ln (| 1 |) + (- 5^{- 1}) + 4^{- 1}$

Step 8.3

Simplify.

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Step 8.3.1

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

$\ln (| 2 |) - \ln (| 1 |) - \frac{1}{5} + 4^{- 1}$

Step 8.3.2

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

$\ln (| 2 |) - \ln (| 1 |) - \frac{1}{5} + \frac{1}{4}$

Step 8.3.3

To write $- \frac{1}{5}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\ln (| 2 |) - \ln (| 1 |) - \frac{1}{5} \cdot \frac{4}{4} + \frac{1}{4}$

Step 8.3.4

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\ln (| 2 |) - \ln (| 1 |) - \frac{1}{5} \cdot \frac{4}{4} + \frac{1}{4} \cdot \frac{5}{5}$

Step 8.3.5

Write each expression with a common denominator of $20$ , by multiplying each by an appropriate factor of $1$ .

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Step 8.3.5.1

Multiply $\frac{1}{5}$ by $\frac{4}{4}$ .

$\ln (| 2 |) - \ln (| 1 |) - \frac{4}{5 \cdot 4} + \frac{1}{4} \cdot \frac{5}{5}$

Step 8.3.5.2

Multiply $5$ by $4$ .

$\ln (| 2 |) - \ln (| 1 |) - \frac{4}{20} + \frac{1}{4} \cdot \frac{5}{5}$

Step 8.3.5.3

Multiply $\frac{1}{4}$ by $\frac{5}{5}$ .

$\ln (| 2 |) - \ln (| 1 |) - \frac{4}{20} + \frac{5}{4 \cdot 5}$

Step 8.3.5.4

Multiply $4$ by $5$ .

$\ln (| 2 |) - \ln (| 1 |) - \frac{4}{20} + \frac{5}{20}$

Step 8.3.6

Combine the numerators over the common denominator.

$\ln (| 2 |) - \ln (| 1 |) + \frac{- 4 + 5}{20}$

Step 8.3.7

Add $- 4$ and $5$ .

$\ln (| 2 |) - \ln (| 1 |) + \frac{1}{20}$

Step 9

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

$\ln (\frac{| 2 |}{| 1 |}) + \frac{1}{20}$

Step 10

Simplify.

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Step 10.1

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

$\ln (\frac{2}{| 1 |}) + \frac{1}{20}$

Step 10.2

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

$\ln (\frac{2}{1}) + \frac{1}{20}$

Step 10.3

Divide $2$ by $1$ .

$\ln (2) + \frac{1}{20}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\ln (2) + \frac{1}{20}$

Decimal Form:

$0.74314718 \dots$

Step 12