Calculus Examples

$\ln (x^{2} - x)$

Step 1

Write $\ln (x^{2} - x)$ as a function.

$f (x) = \ln (x^{2} - x)$

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.

$F (x) = \int \ln (x^{2} - x) d x$

Step 4

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x^{2} - x)$ and $d v = 1$ .

$\ln (x^{2} - x) x - \int x \frac{2 x - 1}{x (x - 1)} d x$

Step 5

Simplify.

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Step 5.1

Combine $x$ and $\frac{2 x - 1}{x (x - 1)}$ .

$\ln (x^{2} - x) x - \int \frac{x (2 x - 1)}{x (x - 1)} d x$

Step 5.2

Cancel the common factor of $x$ .

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Step 5.2.1

Cancel the common factor.

$\ln (x^{2} - x) x - \int \frac{x (2 x - 1)}{x (x - 1)} d x$

Step 5.2.2

Rewrite the expression.

$\ln (x^{2} - x) x - \int \frac{2 x - 1}{x - 1} d x$

Step 6

Divide $2 x - 1$ by $x - 1$ .

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Step 6.1

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

$x$

$1$

$2 x$

$1$

Step 6.2

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $x$ .

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

Step 6.3

Multiply the new quotient term by the divisor.

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

Step 6.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x - 2$

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

Step 6.5

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

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

Step 6.6

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

$\ln (x^{2} - x) x - \int 2 + \frac{1}{x - 1} d x$

Step 7

Split the single integral into multiple integrals.

$\ln (x^{2} - x) x - (\int 2 d x + \int \frac{1}{x - 1} d x)$

Step 8

Apply the constant rule.

$\ln (x^{2} - x) x - (2 x + C + \int \frac{1}{x - 1} d x)$

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $x - 1$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

$\ln (x^{2} - x) x - (2 x + C + \int \frac{1}{u} d u)$

Step 10

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

$\ln (x^{2} - x) x - (2 x + C + \ln (| u |) + C)$

Step 11

Simplify.

$\ln (x^{2} - x) x - 2 x - \ln (| u |) + C$

Step 12

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

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

Step 13

The answer is the antiderivative of the function $f (x) = \ln (x^{2} - x)$ .

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