Calculus Examples

$\frac{x + 1}{x - 1}$

Step 1

Write $\frac{x + 1}{x - 1}$ as a function.

$f (x) = \frac{x + 1}{x - 1}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \frac{x + 1}{x - 1} d x$

Step 4

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

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Step 4.1

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 4.2

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

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

Step 4.3

Multiply the new quotient term by the divisor.

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

Step 4.4

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 4.5

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 4.6

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

$\int 1 + \frac{2}{x - 1} d x$

Step 5

Split the single integral into multiple integrals.

$\int d x + \int \frac{2}{x - 1} d x$

Step 6

Apply the constant rule.

$x + C + \int \frac{2}{x - 1} d x$

Step 7

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

$x + C + 2 \int \frac{1}{x - 1} d x$

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $x - 1$ .

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

Step 8.1.2

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 8.1.3

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 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

Rewrite the problem using $u$ and $d u$ .

$x + C + 2 \int \frac{1}{u} d u$

Step 9

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

$x + C + 2 (\ln (| u |) + C)$

Step 10

Simplify.

$x + 2 \ln (| u |) + C$

Step 11

Replace all occurrences of $u$ with $x - 1$ .

$x + 2 \ln (| x - 1 |) + C$

Step 12

The answer is the antiderivative of the function $f (x) = \frac{x + 1}{x - 1}$ .

$F (x) =$ $x + 2 \ln (| x - 1 |) + C$