Calculus Examples

$\int_{0}^{1} \frac{5 x + 4}{2 x + 1} d x$

Step 1

Let $u = 2 x + 1$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = 2 x + 1$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 1.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 1.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 1.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 1.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 1.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

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

$2 + 0$

Step 1.1.4.2

Add $2$ and $0$ .

$2$

Step 1.2

Substitute the lower limit in for $x$ in $u = 2 x + 1$ .

$u_{lower} = 2 \cdot 0 + 1$

Step 1.3

Simplify.

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Step 1.3.1

Multiply $2$ by $0$ .

$u_{lower} = 0 + 1$

Step 1.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 1.4

Substitute the upper limit in for $x$ in $u = 2 x + 1$ .

$u_{upper} = 2 \cdot 1 + 1$

Step 1.5

Simplify.

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Step 1.5.1

Multiply $2$ by $1$ .

$u_{upper} = 2 + 1$

Step 1.5.2

Add $2$ and $1$ .

$u_{upper} = 3$

Step 1.6

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

$u_{lower} = 1$

$u_{upper} = 3$

Step 1.7

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

$\int_{1}^{3} \frac{5 (\frac{u}{2} - \frac{1}{2}) + 4}{u} \cdot \frac{1}{2} d u$

Step 2

Simplify.

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Step 2.1

Multiply $\frac{5 (\frac{u}{2} - \frac{1}{2}) + 4}{u}$ by $\frac{1}{2}$ .

$\int_{1}^{3} \frac{5 (\frac{u}{2} - \frac{1}{2}) + 4}{u \cdot 2} d u$

Step 2.2

Move $2$ to the left of $u$ .

$\int_{1}^{3} \frac{5 (\frac{u}{2} - \frac{1}{2}) + 4}{2 u} d u$

Step 3

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int_{1}^{3} \frac{5 (\frac{u}{2} - \frac{1}{2}) + 4}{u} d u$

Step 4

Split the fraction into multiple fractions.

$\frac{1}{2} \int_{1}^{3} \frac{5 (\frac{u}{2} - \frac{1}{2})}{u} + \frac{4}{u} d u$

Step 5

Split the single integral into multiple integrals.

$\frac{1}{2} (\int_{1}^{3} \frac{5 (\frac{u}{2} - \frac{1}{2})}{u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 6

Since $5$ is constant with respect to $u$ , move $5$ out of the integral.

$\frac{1}{2} (5 \int_{1}^{3} \frac{\frac{u}{2} - \frac{1}{2}}{u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{\frac{u}{2} - \frac{1}{2}}{u}$ by $\frac{2}{2}$ .

$\frac{1}{2} (5 \int_{1}^{3} \frac{2}{2} \cdot \frac{\frac{u}{2} - \frac{1}{2}}{u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.2

Combine.

$\frac{1}{2} (5 \int_{1}^{3} \frac{2 (\frac{u}{2} - \frac{1}{2})}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.3

Apply the distributive property.

$\frac{1}{2} (5 \int_{1}^{3} \frac{2 \frac{u}{2} + 2 (- \frac{1}{2})}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.4

Cancel the common factor of $2$ .

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Step 7.4.1

Cancel the common factor.

$\frac{1}{2} (5 \int_{1}^{3} \frac{2 \frac{u}{2} + 2 (- \frac{1}{2})}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.4.2

Rewrite the expression.

$\frac{1}{2} (5 \int_{1}^{3} \frac{u + 2 (- \frac{1}{2})}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.5

Multiply $- 1$ by $2$ .

$\frac{1}{2} (5 \int_{1}^{3} \frac{u - 2 (\frac{1}{2})}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.6

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

$\frac{1}{2} (5 \int_{1}^{3} \frac{u + \frac{- 2}{2}}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.7

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

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Step 7.7.1

Factor $2$ out of $- 2$ .

$\frac{1}{2} (5 \int_{1}^{3} \frac{u + \frac{2 \cdot - 1}{2}}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.7.2

Cancel the common factors.

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Step 7.7.2.1

Factor $2$ out of $2$ .

$\frac{1}{2} (5 \int_{1}^{3} \frac{u + \frac{2 \cdot - 1}{2 (1)}}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.7.2.2

Cancel the common factor.

$\frac{1}{2} (5 \int_{1}^{3} \frac{u + \frac{2 \cdot - 1}{2 \cdot 1}}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.7.2.3

Rewrite the expression.

$\frac{1}{2} (5 \int_{1}^{3} \frac{u + \frac{- 1}{1}}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 7.7.2.4

Divide $- 1$ by $1$ .

$\frac{1}{2} (5 \int_{1}^{3} \frac{u - 1}{2 u} d u + \int_{1}^{3} \frac{4}{u} d u)$

Step 8

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} (5 (\frac{1}{2} \int_{1}^{3} \frac{u - 1}{u} d u) + \int_{1}^{3} \frac{4}{u} d u)$

Step 9

Split the fraction into multiple fractions.

$\frac{1}{2} (5 (\frac{1}{2} \int_{1}^{3} \frac{u}{u} + \frac{- 1}{u} d u) + \int_{1}^{3} \frac{4}{u} d u)$

Step 10

Split the single integral into multiple integrals.

$\frac{1}{2} (5 (\frac{1}{2} (\int_{1}^{3} \frac{u}{u} d u + \int_{1}^{3} \frac{- 1}{u} d u)) + \int_{1}^{3} \frac{4}{u} d u)$

Step 11

Cancel the common factor of $u$ .

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Step 11.1

Cancel the common factor.

$\frac{1}{2} (5 (\frac{1}{2} (\int_{1}^{3} \frac{u}{u} d u + \int_{1}^{3} \frac{- 1}{u} d u)) + \int_{1}^{3} \frac{4}{u} d u)$

Step 11.2

Rewrite the expression.

$\frac{1}{2} (5 (\frac{1}{2} (\int_{1}^{3} d u + \int_{1}^{3} \frac{- 1}{u} d u)) + \int_{1}^{3} \frac{4}{u} d u)$

Step 12

Apply the constant rule.

$\frac{1}{2} (5 (\frac{1}{2} (u]_{1}^{3} + \int_{1}^{3} \frac{- 1}{u} d u)) + \int_{1}^{3} \frac{4}{u} d u)$

Step 13

Move the negative in front of the fraction.

$\frac{1}{2} (5 (\frac{1}{2} (u]_{1}^{3} + \int_{1}^{3} - \frac{1}{u} d u)) + \int_{1}^{3} \frac{4}{u} d u)$

Step 14

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

$\frac{1}{2} (5 (\frac{1}{2} (u]_{1}^{3} - \int_{1}^{3} \frac{1}{u} d u)) + \int_{1}^{3} \frac{4}{u} d u)$

Step 15

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

$\frac{1}{2} (5 (\frac{1}{2} (u]_{1}^{3} - (\ln (| u |)]_{1}^{3}))) + \int_{1}^{3} \frac{4}{u} d u)$

Step 16

Combine $\frac{1}{2}$ and $5$ .

$\frac{1}{2} (\frac{5}{2} (u]_{1}^{3} - (\ln (| u |)]_{1}^{3})) + \int_{1}^{3} \frac{4}{u} d u)$

Step 17

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

$\frac{1}{2} (\frac{5}{2} (u]_{1}^{3} - (\ln (| u |)]_{1}^{3})) + 4 \int_{1}^{3} \frac{1}{u} d u)$

Step 18

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

$\frac{1}{2} (\frac{5}{2} (u]_{1}^{3} - (\ln (| u |)]_{1}^{3})) + 4 (\ln (| u |)]_{1}^{3}))$

Step 19

Replace all occurrences of $u$ with $2 x + 1$ .

$\frac{1}{2} (\frac{5}{2} (2 x + 1]_{1}^{3} - (\ln (| 2 x + 1 |)]_{1}^{3})) + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20

Simplify.

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Step 20.1

Apply the distributive property.

$\frac{1}{2} (\frac{5}{2} 2 x + 1]_{1}^{3} + \frac{5}{2} (- (\ln (| 2 x + 1 |)]_{1}^{3})) + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20.2

Combine $\frac{5}{2}$ and $2 x + 1]_{1}^{3}$ .

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3})}{2} + \frac{5}{2} (- (\ln (| 2 x + 1 |)]_{1}^{3})) + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20.3

Combine $\frac{5}{2}$ and $\ln (| 2 x + 1 |)]_{1}^{3}$ .

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3})}{2} - \frac{5 (\ln (| 2 x + 1 |)]_{1}^{3})}{2} + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20.4

Combine the numerators over the common denominator.

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3}) - 5 (\ln (| 2 x + 1 |)]_{1}^{3})}{2} + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20.5

Factor $5$ out of $5 (2 x + 1]_{1}^{3}) - 5 (\ln (| 2 x + 1 |)]_{1}^{3})$ .

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Step 20.5.1

Factor $5$ out of $- 5 (\ln (| 2 x + 1 |)]_{1}^{3})$ .

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3}) + 5 (- (\ln (| 2 x + 1 |)]_{1}^{3}))}{2} + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20.5.2

Factor $5$ out of $5 (2 x + 1]_{1}^{3}) + 5 (- (\ln (| 2 x + 1 |)]_{1}^{3}))$ .

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3} - (\ln (| 2 x + 1 |)]_{1}^{3}))}{2} + 4 (\ln (| 2 x + 1 |)]_{1}^{3}))$

Step 20.6

To write $4 (\ln (| 2 x + 1 |)]_{1}^{3})$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3} - (\ln (| 2 x + 1 |)]_{1}^{3}))}{2} + 4 (\ln (| 2 x + 1 |)]_{1}^{3}) \cdot \frac{2}{2})$

Step 20.7

Combine $4 (\ln (| 2 x + 1 |)]_{1}^{3})$ and $\frac{2}{2}$ .

$\frac{1}{2} (\frac{5 (2 x + 1]_{1}^{3} - (\ln (| 2 x + 1 |)]_{1}^{3}))}{2} + \frac{4 (\ln (| 2 x + 1 |)]_{1}^{3}) \cdot 2}{2})$

Step 20.8

Combine the numerators over the common denominator.

$\frac{1}{2} \cdot \frac{5 (2 x + 1]_{1}^{3} - (\ln (| 2 x + 1 |)]_{1}^{3})) + 4 (\ln (| 2 x + 1 |)]_{1}^{3}) \cdot 2}{2}$

Step 20.9

Simplify the numerator.

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Step 20.9.1

Apply the distributive property.

$\frac{1}{2} \cdot \frac{5 (2 x + 1]_{1}^{3}) + 5 (- (\ln (| 2 x + 1 |)]_{1}^{3})) + 4 (\ln (| 2 x + 1 |)]_{1}^{3}) \cdot 2}{2}$

Step 20.9.2

Multiply $- 1$ by $5$ .

$\frac{1}{2} \cdot \frac{5 (2 x + 1]_{1}^{3}) - 5 (\ln (| 2 x + 1 |)]_{1}^{3}) + 4 (\ln (| 2 x + 1 |)]_{1}^{3}) \cdot 2}{2}$

Step 20.9.3

Multiply $2$ by $4$ .

$\frac{1}{2} \cdot \frac{5 (2 x + 1]_{1}^{3}) - 5 (\ln (| 2 x + 1 |)]_{1}^{3}) + 8 (\ln (| 2 x + 1 |)]_{1}^{3})}{2}$

Step 20.9.4

Add $- 5 (\ln (| 2 x + 1 |)]_{1}^{3})$ and $8 (\ln (| 2 x + 1 |)]_{1}^{3})$ .

$\frac{1}{2} \cdot \frac{5 (2 x + 1]_{1}^{3}) + 3 (\ln (| 2 x + 1 |)]_{1}^{3})}{2}$

Step 20.10

Multiply $\frac{1}{2} \cdot \frac{5 (2 x + 1]_{1}^{3}) + 3 (\ln (| 2 x + 1 |)]_{1}^{3})}{2}$ .

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Step 20.10.1

Multiply $\frac{1}{2}$ by $\frac{5 (2 x + 1]_{1}^{3}) + 3 (\ln (| 2 x + 1 |)]_{1}^{3})}{2}$ .

$\frac{5 (2 x + 1]_{1}^{3}) + 3 (\ln (| 2 x + 1 |)]_{1}^{3})}{2 \cdot 2}$

Step 20.10.2

Multiply $2$ by $2$ .

$\frac{5 (2 x + 1]_{1}^{3}) + 3 (\ln (| 2 x + 1 |)]_{1}^{3})}{4}$

Step 21

Reorder terms.

$\frac{1}{4} (5 (2 x + 1]_{1}^{3}) + 3 (\ln (| 2 x + 1 |)]_{1}^{3}))$