Calculus Examples

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

Step 1

Divide $x - 1$ by $x + 1$ .

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x$

$1$

Step 1.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

					$1$
	$x$	+	$1$		$x$	-	$1$

Step 1.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$1$		$x$	-	$1$
			+	$x$	+	$1$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 1$

				$1$
$x$	+	$1$		$x$	-	$1$
			-	$x$	-	$1$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$x$	+	$1$		$x$	-	$1$
			-	$x$	-	$1$
					-	$2$

Step 1.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

Apply the constant rule.

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

Step 4

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 5

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 6

Multiply $2$ by $- 1$ .

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

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x + 1$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$u_{lower} = 0 + 1$

Step 7.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 7.4

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

$u_{upper} = 1 + 1$

Step 7.5

Add $1$ and $1$ .

$u_{upper} = 2$

Step 7.6

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

$u_{lower} = 1$

$u_{upper} = 2$

Step 7.7

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

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

Step 8

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

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

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $x$ at $1$ and at $0$ .

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

Step 9.2

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

$(1) + 0 - 2 ((\ln (| 2 |)) - \ln (| 1 |))$

Step 9.3

Add $1$ and $0$ .

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

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

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

Step 11.2

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

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

Step 11.3

Divide $2$ by $1$ .

$1 - 2 \ln (2)$

Step 12

The result can be shown in multiple forms.

Exact Form:

$1 - 2 \ln (2)$

Decimal Form:

$- 0.38629436 \dots$

Step 13