Calculus Examples

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

Step 1

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

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

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Step 1.1.1

Differentiate $x + 1$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

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

$1 + 0$

Step 1.1.5

Add $1$ and $0$ .

$1$

Step 1.2

Substitute the lower limit in for $x$ in $u = x + 1$ .

$u_{lower} = 0 + 1$

Step 1.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 1.4

Substitute the upper limit in for $x$ in $u = x + 1$ .

$u_{upper} = 1 + 1$

Step 1.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 1.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 1.7

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

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

Step 2

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

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

Step 3

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

$\ln (| 2 |) - \ln (| 1 |)$

Step 4

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

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

Step 5

Simplify.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Divide $2$ by $1$ .

$\ln (2)$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\ln (2)$

Decimal Form:

$0.69314718 \dots$

Step 7