Calculus Examples

$\int_{1}^{3} \frac{6 x^{2}}{8 x^{3} + 1} d x$

Step 1

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int_{1}^{3} \frac{x^{2}}{8 x^{3} + 1} d x$

Step 2

Rewrite $x^{3}$ as ${(x^{3})}^{1}$ .

$6 \int_{1}^{3} \frac{x^{2}}{8 {(x^{3})}^{1} + 1} d x$

Step 3

Let $u_{1} = x^{3}$ . Then $d u_{1} = 3 x^{2} d x$ , so $\frac{1}{3} d u_{1} = x^{2} d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 3.1

Let $u_{1} = x^{3}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $x^{3}$ .

$\frac{d}{d x} [x^{3}]$

Step 3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2}$

Step 3.2

Substitute the lower limit in for $x$ in $u_{1} = x^{3}$ .

$u_{lower} = 1^{3}$

Step 3.3

One to any power is one.

$u_{lower} = 1$

Step 3.4

Substitute the upper limit in for $x$ in $u_{1} = x^{3}$ .

$u_{upper} = 3^{3}$

Step 3.5

Raise $3$ to the power of $3$ .

$u_{upper} = 27$

Step 3.6

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

$u_{lower} = 1$

$u_{upper} = 27$

Step 3.7

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

$6 \int_{1}^{27} \frac{1}{8 {u_{1}}^{1} + 1} \cdot \frac{1}{3} d u_{1}$

Step 4

Simplify.

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Step 4.1

Simplify.

$6 \int_{1}^{27} \frac{1}{8 u_{1} + 1} \cdot \frac{1}{3} d u_{1}$

Step 4.2

Multiply $\frac{1}{8 u_{1} + 1}$ by $\frac{1}{3}$ .

$6 \int_{1}^{27} \frac{1}{(8 u_{1} + 1) \cdot 3} d u_{1}$

Step 4.3

Move $3$ to the left of $8 u_{1} + 1$ .

$6 \int_{1}^{27} \frac{1}{3 (8 u_{1} + 1)} d u_{1}$

Step 5

Since $\frac{1}{3}$ is constant with respect to $u_{1}$ , move $\frac{1}{3}$ out of the integral.

$6 (\frac{1}{3} \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1})$

Step 6

Simplify.

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Step 6.1

Combine $\frac{1}{3}$ and $6$ .

$\frac{6}{3} \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1}$

Step 6.2

Cancel the common factor of $6$ and $3$ .

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Step 6.2.1

Factor $3$ out of $6$ .

$\frac{3 \cdot 2}{3} \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1}$

Step 6.2.2

Cancel the common factors.

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Step 6.2.2.1

Factor $3$ out of $3$ .

$\frac{3 \cdot 2}{3 (1)} \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1}$

Step 6.2.2.2

Cancel the common factor.

$\frac{3 \cdot 2}{3 \cdot 1} \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1}$

Step 6.2.2.3

Rewrite the expression.

$\frac{2}{1} \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1}$

Step 6.2.2.4

Divide $2$ by $1$ .

$2 \int_{1}^{27} \frac{1}{8 u_{1} + 1} d u_{1}$

Step 7

Let $u_{2} = 8 u_{1} + 1$ . Then $d u_{2} = 8 d u_{1}$ , so $\frac{1}{8} d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 7.1

Let $u_{2} = 8 u_{1} + 1$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 7.1.1

Differentiate $8 u_{1} + 1$ .

$\frac{d}{d u_{1}} [8 u_{1} + 1]$

Step 7.1.2

By the Sum Rule, the derivative of $8 u_{1} + 1$ with respect to $u_{1}$ is $\frac{d}{d u_{1}} [8 u_{1}] + \frac{d}{d u_{1}} [1]$ .

$\frac{d}{d u_{1}} [8 u_{1}] + \frac{d}{d u_{1}} [1]$

Step 7.1.3

Evaluate $\frac{d}{d u_{1}} [8 u_{1}]$ .

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Step 7.1.3.1

Since $8$ is constant with respect to $u_{1}$ , the derivative of $8 u_{1}$ with respect to $u_{1}$ is $8 \frac{d}{d u_{1}} [u_{1}]$ .

$8 \frac{d}{d u_{1}} [u_{1}] + \frac{d}{d u_{1}} [1]$

Step 7.1.3.2

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

$8 \cdot 1 + \frac{d}{d u_{1}} [1]$

Step 7.1.3.3

Multiply $8$ by $1$ .

$8 + \frac{d}{d u_{1}} [1]$

Step 7.1.4

Differentiate using the Constant Rule.

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Step 7.1.4.1

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$8 + 0$

Step 7.1.4.2

Add $8$ and $0$ .

$8$

Step 7.2

Substitute the lower limit in for $u_{1}$ in $u_{2} = 8 u_{1} + 1$ .

$u_{lower} = 8 \cdot 1 + 1$

Step 7.3

Simplify.

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Step 7.3.1

Multiply $8$ by $1$ .

$u_{lower} = 8 + 1$

Step 7.3.2

Add $8$ and $1$ .

$u_{lower} = 9$

Step 7.4

Substitute the upper limit in for $u_{1}$ in $u_{2} = 8 u_{1} + 1$ .

$u_{upper} = 8 \cdot 27 + 1$

Step 7.5

Simplify.

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Step 7.5.1

Multiply $8$ by $27$ .

$u_{upper} = 216 + 1$

Step 7.5.2

Add $216$ and $1$ .

$u_{upper} = 217$

Step 7.6

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

$u_{lower} = 9$

$u_{upper} = 217$

Step 7.7

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

$2 \int_{9}^{217} \frac{1}{u_{2}} \cdot \frac{1}{8} d u_{2}$

Step 8

Simplify.

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Step 8.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{8}$ .

$2 \int_{9}^{217} \frac{1}{u_{2} \cdot 8} d u_{2}$

Step 8.2

Move $8$ to the left of $u_{2}$ .

$2 \int_{9}^{217} \frac{1}{8 u_{2}} d u_{2}$

Step 9

Since $\frac{1}{8}$ is constant with respect to $u_{2}$ , move $\frac{1}{8}$ out of the integral.

$2 (\frac{1}{8} \int_{9}^{217} \frac{1}{u_{2}} d u_{2})$

Step 10

Simplify.

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Step 10.1

Combine $\frac{1}{8}$ and $2$ .

$\frac{2}{8} \int_{9}^{217} \frac{1}{u_{2}} d u_{2}$

Step 10.2

Cancel the common factor of $2$ and $8$ .

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Step 10.2.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{8} \int_{9}^{217} \frac{1}{u_{2}} d u_{2}$

Step 10.2.2

Cancel the common factors.

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Step 10.2.2.1

Factor $2$ out of $8$ .

$\frac{2 \cdot 1}{2 \cdot 4} \int_{9}^{217} \frac{1}{u_{2}} d u_{2}$

Step 10.2.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 4} \int_{9}^{217} \frac{1}{u_{2}} d u_{2}$

Step 10.2.2.3

Rewrite the expression.

$\frac{1}{4} \int_{9}^{217} \frac{1}{u_{2}} d u_{2}$

Step 11

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{4} \ln (| u_{2} |)]_{9}^{217}$

Step 12

Evaluate $\ln (| u_{2} |)$ at $217$ and at $9$ .

$\frac{1}{4} (\ln (| 217 |) - \ln (| 9 |))$

Step 13

Simplify.

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Step 13.1

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

$\frac{1}{4} \ln (\frac{| 217 |}{| 9 |})$

Step 13.2

Combine $\frac{1}{4}$ and $\ln (\frac{| 217 |}{| 9 |})$ .

$\frac{\ln (\frac{| 217 |}{| 9 |})}{4}$

Step 14

Simplify.

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Step 14.1

The absolute value is the distance between a number and zero. The distance between $0$ and $217$ is $217$ .

$\frac{\ln (\frac{217}{| 9 |})}{4}$

Step 14.2

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

$\frac{\ln (\frac{217}{9})}{4}$

Step 15

The result can be shown in multiple forms.

Exact Form:

$\frac{\ln (\frac{217}{9})}{4}$

Decimal Form:

$0.79566819 \dots$

Step 16