Calculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Let $u = x^{2} + 1$ . 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 3.1

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

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Step 3.1.1

Differentiate $x^{2} + 1$ .

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

Step 3.1.2

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

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

Step 3.1.3

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

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

Step 3.1.4

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

$2 x + 0$

Step 3.1.5

Add $2 x$ and $0$ .

$2 x$

Step 3.2

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

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

Step 3.3

Simplify.

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Step 3.3.1

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

$u_{lower} = 0 + 1$

Step 3.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 3.4

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

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

Step 3.5

Simplify.

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Step 3.5.1

One to any power is one.

$u_{upper} = 1 + 1$

Step 3.5.2

Add $1$ and $1$ .

$u_{upper} = 2$

Step 3.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 3.7

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

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

Step 4

Simplify.

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Step 4.1

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

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

Step 4.2

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

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

Step 5

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

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

Step 6

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

$\arctan (x) x]_{0}^{1} - \frac{1}{2} \ln (| u |)]_{1}^{2}$

Step 7

Substitute and simplify.

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Step 7.1

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

$(\arctan (1) \cdot 1) - \arctan (0) \cdot 0 - \frac{1}{2} \ln (| u |)]_{1}^{2}$

Step 7.2

Evaluate $\ln (| u |)$ at $2$ and at $1$ .

$(\arctan (1) \cdot 1) - \arctan (0) \cdot 0 - \frac{1}{2} ((\ln (| 2 |)) - \ln (| 1 |))$

Step 7.3

Simplify.

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Step 7.3.1

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

$\arctan (1) - \arctan (0) \cdot 0 - \frac{1}{2} ((\ln (| 2 |)) - \ln (| 1 |))$

Step 7.3.2

Multiply $0$ by $- 1$ .

$\arctan (1) + 0 \arctan (0) - \frac{1}{2} ((\ln (| 2 |)) - \ln (| 1 |))$

Step 7.3.3

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

$\arctan (1) + 0 - \frac{1}{2} ((\ln (| 2 |)) - \ln (| 1 |))$

Step 7.3.4

Add $\arctan (1)$ and $0$ .

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

Step 8

Simplify.

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Step 8.1

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

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

Step 8.2

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

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

Step 9

Simplify.

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Step 9.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Step 9.2

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

$\arctan (1) - \frac{\ln (\frac{2}{1})}{2}$

Step 9.3

Divide $2$ by $1$ .

$\arctan (1) - \frac{\ln (2)}{2}$

Step 10

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$\frac{π}{4} - \frac{\ln (2)}{2}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\frac{π}{4} - \frac{\ln (2)}{2}$

Decimal Form:

$0.43882457 \dots$