Calculus Examples

$\int_{0}^{1} \arctan (4 x) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \arctan (4 x)$ and $d v = 1$ .

$\arctan (4 x) x]_{0}^{1} - \int_{0}^{1} x \frac{4}{16 x^{2} + 1} d x$

Step 2

Simplify.

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Step 2.1

Combine $x$ and $\frac{4}{16 x^{2} + 1}$ .

$\arctan (4 x) x]_{0}^{1} - \int_{0}^{1} \frac{x \cdot 4}{16 x^{2} + 1} d x$

Step 2.2

Move $4$ to the left of $x$ .

$\arctan (4 x) x]_{0}^{1} - \int_{0}^{1} \frac{4 x}{16 x^{2} + 1} d x$

Step 3

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

$\arctan (4 x) x]_{0}^{1} - (4 \int_{0}^{1} \frac{x}{16 x^{2} + 1} d x)$

Step 4

Multiply $4$ by $- 1$ .

$\arctan (4 x) x]_{0}^{1} - 4 \int_{0}^{1} \frac{x}{16 x^{2} + 1} d x$

Step 5

Let $u = 16 x^{2} + 1$ . Then $d u = 32 x d x$ , so $\frac{1}{32} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = 16 x^{2} + 1$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $16 x^{2} + 1$ .

$\frac{d}{d x} [16 x^{2} + 1]$

Step 5.1.2

By the Sum Rule, the derivative of $16 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [1]$

Step 5.1.3

Evaluate $\frac{d}{d x} [16 x^{2}]$ .

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Step 5.1.3.1

Since $16$ is constant with respect to $x$ , the derivative of $16 x^{2}$ with respect to $x$ is $16 \frac{d}{d x} [x^{2}]$ .

$16 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 5.1.3.2

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

$16 (2 x) + \frac{d}{d x} [1]$

Step 5.1.3.3

Multiply $2$ by $16$ .

$32 x + \frac{d}{d x} [1]$

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

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

$32 x + 0$

Step 5.1.4.2

Add $32 x$ and $0$ .

$32 x$

Step 5.2

Substitute the lower limit in for $x$ in $u = 16 x^{2} + 1$ .

$u_{lower} = 16 \cdot 0^{2} + 1$

Step 5.3

Simplify.

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Step 5.3.1

Simplify each term.

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Step 5.3.1.1

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

$u_{lower} = 16 \cdot 0 + 1$

Step 5.3.1.2

Multiply $16$ by $0$ .

$u_{lower} = 0 + 1$

Step 5.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 5.4

Substitute the upper limit in for $x$ in $u = 16 x^{2} + 1$ .

$u_{upper} = 16 \cdot 1^{2} + 1$

Step 5.5

Simplify.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

One to any power is one.

$u_{upper} = 16 \cdot 1 + 1$

Step 5.5.1.2

Multiply $16$ by $1$ .

$u_{upper} = 16 + 1$

Step 5.5.2

Add $16$ and $1$ .

$u_{upper} = 17$

Step 5.6

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

$u_{lower} = 1$

$u_{upper} = 17$

Step 5.7

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

$\arctan (4 x) x]_{0}^{1} - 4 \int_{1}^{17} \frac{1}{u} \cdot \frac{1}{32} d u$

Step 6

Simplify.

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Step 6.1

Multiply $\frac{1}{u}$ by $\frac{1}{32}$ .

$\arctan (4 x) x]_{0}^{1} - 4 \int_{1}^{17} \frac{1}{u \cdot 32} d u$

Step 6.2

Move $32$ to the left of $u$ .

$\arctan (4 x) x]_{0}^{1} - 4 \int_{1}^{17} \frac{1}{32 u} d u$

Step 7

Since $\frac{1}{32}$ is constant with respect to $u$ , move $\frac{1}{32}$ out of the integral.

$\arctan (4 x) x]_{0}^{1} - 4 (\frac{1}{32} \int_{1}^{17} \frac{1}{u} d u)$

Step 8

Simplify.

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Step 8.1

Combine $\frac{1}{32}$ and $- 4$ .

$\arctan (4 x) x]_{0}^{1} + \frac{- 4}{32} \int_{1}^{17} \frac{1}{u} d u$

Step 8.2

Cancel the common factor of $- 4$ and $32$ .

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Step 8.2.1

Factor $4$ out of $- 4$ .

$\arctan (4 x) x]_{0}^{1} + \frac{4 (- 1)}{32} \int_{1}^{17} \frac{1}{u} d u$

Step 8.2.2

Cancel the common factors.

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Step 8.2.2.1

Factor $4$ out of $32$ .

$\arctan (4 x) x]_{0}^{1} + \frac{4 \cdot - 1}{4 \cdot 8} \int_{1}^{17} \frac{1}{u} d u$

Step 8.2.2.2

Cancel the common factor.

$\arctan (4 x) x]_{0}^{1} + \frac{4 \cdot - 1}{4 \cdot 8} \int_{1}^{17} \frac{1}{u} d u$

Step 8.2.2.3

Rewrite the expression.

$\arctan (4 x) x]_{0}^{1} + \frac{- 1}{8} \int_{1}^{17} \frac{1}{u} d u$

Step 8.3

Move the negative in front of the fraction.

$\arctan (4 x) x]_{0}^{1} - \frac{1}{8} \int_{1}^{17} \frac{1}{u} d u$

Step 9

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

$\arctan (4 x) x]_{0}^{1} - \frac{1}{8} \ln (| u |)]_{1}^{17}$

Step 10

Simplify.

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Step 10.1

Evaluate $\arctan (4 x) x$ at $1$ and at $0$ .

$(\arctan (4 \cdot 1) \cdot 1) - \arctan (4 \cdot 0) \cdot 0 - \frac{1}{8} \ln (| u |)]_{1}^{17}$

Step 10.2

Simplify the expression.

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Step 10.2.1

Evaluate $\ln (| u |)$ at $17$ and at $1$ .

$(\arctan (4 \cdot 1) \cdot 1) - \arctan (4 \cdot 0) \cdot 0 - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.2.2

Multiply.

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Step 10.2.2.1

Multiply $4$ by $1$ .

$\arctan (4) \cdot 1 - \arctan (4 \cdot 0) \cdot 0 - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.2.2.2

Multiply $\arctan (4)$ by $1$ .

$\arctan (4) - \arctan (4 \cdot 0) \cdot 0 - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.2.2.3

Multiply $4$ by $0$ .

$\arctan (4) - \arctan (0) \cdot 0 - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.2.3

Simplify.

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Step 10.2.3.1

Multiply $0$ by $- 1$ .

$\arctan (4) + 0 \arctan (0) - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.2.3.2

Multiply $0$ by $\arctan (0)$ .

$\arctan (4) + 0 - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.2.4

Add $\arctan (4)$ and $0$ .

$\arctan (4) - \frac{1}{8} ((\ln (| 17 |)) - \ln (| 1 |))$

Step 10.3

Simplify.

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Step 10.3.1

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

$\arctan (4) - \frac{1}{8} \ln (\frac{| 17 |}{| 1 |})$

Step 10.3.2

Combine $\ln (\frac{| 17 |}{| 1 |})$ and $\frac{1}{8}$ .

$\arctan (4) - \frac{\ln (\frac{| 17 |}{| 1 |})}{8}$

Step 10.4

Simplify.

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Step 10.4.1

The absolute value is the distance between a number and zero. The distance between $0$ and $17$ is $17$ .

$\arctan (4) - \frac{\ln (\frac{17}{| 1 |})}{8}$

Step 10.4.2

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

$\arctan (4) - \frac{\ln (\frac{17}{1})}{8}$

Step 10.4.3

Divide $17$ by $1$ .

$\arctan (4) - \frac{\ln (17)}{8}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\arctan (4) - \frac{\ln (17)}{8}$

Decimal Form:

$0.97166599 \dots$