Calculus Examples

$\int_{0}^{1} \frac{2 x^{3} + x}{x^{2} + x^{4} + 1} d x$

Step 1

Let $u = x^{2} + x^{4} + 1$ . Then $d u = (4 x^{3} + 2 x) d x$ , so $\frac{1}{2} d u = (2 x^{3} + x) d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = x^{2} + x^{4} + 1$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $x^{2} + x^{4} + 1$ .

$\frac{d}{d x} [x^{2} + x^{4} + 1]$

Step 1.1.2

By the Sum Rule, the derivative of $x^{2} + x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$

Step 1.1.3

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

$2 x + \frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$

Step 1.1.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$2 x + 4 x^{3} + \frac{d}{d x} [1]$

Step 1.1.5

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

$2 x + 4 x^{3} + 0$

Step 1.1.6

Simplify.

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Step 1.1.6.1

Add $2 x + 4 x^{3}$ and $0$ .

$2 x + 4 x^{3}$

Step 1.1.6.2

Reorder terms.

$4 x^{3} + 2 x$

Step 1.2

Substitute the lower limit in for $x$ in $u = x^{2} + x^{4} + 1$ .

$u_{lower} = 0^{2} + 0^{4} + 1$

Step 1.3

Simplify.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

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

$u_{lower} = 0 + 0^{4} + 1$

Step 1.3.1.2

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

$u_{lower} = 0 + 0 + 1$

Step 1.3.2

Add $0$ and $0$ .

$u_{lower} = 0 + 1$

Step 1.3.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 1.4

Substitute the upper limit in for $x$ in $u = x^{2} + x^{4} + 1$ .

$u_{upper} = 1^{2} + 1^{4} + 1$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

One to any power is one.

$u_{upper} = 1 + 1^{4} + 1$

Step 1.5.1.2

One to any power is one.

$u_{upper} = 1 + 1 + 1$

Step 1.5.2

Add $1$ and $1$ .

$u_{upper} = 2 + 1$

Step 1.5.3

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.

$\int_{1}^{3} \frac{1}{u} \cdot \frac{1}{2} d u$

Step 2

Simplify.

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Step 2.1

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

$\int_{1}^{3} \frac{1}{u \cdot 2} d u$

Step 2.2

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

$\int_{1}^{3} \frac{1}{2 u} d u$

Step 3

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

$\frac{1}{2} \int_{1}^{3} \frac{1}{u} d u$

Step 4

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

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

Step 5

Evaluate $\ln (| u |)$ at $3$ and at $1$ .

$\frac{1}{2} (\ln (| 3 |) - \ln (| 1 |))$

Step 6

Simplify.

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Step 6.1

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

$\frac{1}{2} \ln (\frac{| 3 |}{| 1 |})$

Step 6.2

Combine $\frac{1}{2}$ and $\ln (\frac{| 3 |}{| 1 |})$ .

$\frac{\ln (\frac{| 3 |}{| 1 |})}{2}$

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 $3$ is $3$ .

$\frac{\ln (\frac{3}{| 1 |})}{2}$

Step 7.2

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

$\frac{\ln (\frac{3}{1})}{2}$

Step 7.3

Divide $3$ by $1$ .

$\frac{\ln (3)}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\ln (3)}{2}$

Decimal Form:

$0.54930614 \dots$

Step 9