Calculus Examples

$\int_{0}^{5} \frac{x}{x^{2} + 10} d x$

Step 1

Let $u = x^{2} + 10$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = x^{2} + 10$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $x^{2} + 10$ .

$\frac{d}{d x} [x^{2} + 10]$

Step 1.1.2

By the Sum Rule, the derivative of $x^{2} + 10$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [10]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [10]$

Step 1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [10]$

Step 1.1.4

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$2 x + 0$

Step 1.1.5

Add $2 x$ and $0$ .

$2 x$

Step 1.2

Substitute the lower limit in for $x$ in $u = x^{2} + 10$ .

$u_{lower} = 0^{2} + 10$

Step 1.3

Simplify.

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Step 1.3.1

Raising $0$ to any positive power yields $0$ .

$u_{lower} = 0 + 10$

Step 1.3.2

Add $0$ and $10$ .

$u_{lower} = 10$

Step 1.4

Substitute the upper limit in for $x$ in $u = x^{2} + 10$ .

$u_{upper} = 5^{2} + 10$

Step 1.5

Simplify.

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Step 1.5.1

Raise $5$ to the power of $2$ .

$u_{upper} = 25 + 10$

Step 1.5.2

Add $25$ and $10$ .

$u_{upper} = 35$

Step 1.6

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

$u_{lower} = 10$

$u_{upper} = 35$

Step 1.7

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

$\int_{10}^{35} \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2

Simplify.

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Step 2.1

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

$\int_{10}^{35} \frac{1}{u \cdot 2} d u$

Step 2.2

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

$\int_{10}^{35} \frac{1}{2 u} d u$

Step 3

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

$\frac{1}{2} \int_{10}^{35} \frac{1}{u} d u$

Step 4

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

$\frac{1}{2} \ln (| u |)]_{10}^{35}$

Step 5

Evaluate $\ln (| u |)$ at $35$ and at $10$ .

$\frac{1}{2} (\ln (| 35 |) - \ln (| 10 |))$

Step 6

Simplify.

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Step 6.1

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

$\frac{1}{2} \ln (\frac{| 35 |}{| 10 |})$

Step 6.2

Combine $\frac{1}{2}$ and $\ln (\frac{| 35 |}{| 10 |})$ .

$\frac{\ln (\frac{| 35 |}{| 10 |})}{2}$

Step 7

Simplify.

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Step 7.1

The absolute value is the distance between a number and zero. The distance between $0$ and $35$ is $35$ .

$\frac{\ln (\frac{35}{| 10 |})}{2}$

Step 7.2

The absolute value is the distance between a number and zero. The distance between $0$ and $10$ is $10$ .

$\frac{\ln (\frac{35}{10})}{2}$

Step 7.3

Cancel the common factor of $35$ and $10$ .

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Step 7.3.1

Factor $5$ out of $35$ .

$\frac{\ln (\frac{5 (7)}{10})}{2}$

Step 7.3.2

Cancel the common factors.

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Step 7.3.2.1

Factor $5$ out of $10$ .

$\frac{\ln (\frac{5 \cdot 7}{5 \cdot 2})}{2}$

Step 7.3.2.2

Cancel the common factor.

$\frac{\ln (\frac{5 \cdot 7}{5 \cdot 2})}{2}$

Step 7.3.2.3

Rewrite the expression.

$\frac{\ln (\frac{7}{2})}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\ln (\frac{7}{2})}{2}$

Decimal Form:

$0.62638148 \dots$

Step 9